Vector JEE Mathematics Worksheets Set 02

Question. If \(\vec{A}, \vec{B} \, \& \, \vec{C}\) are vectors such that \(|\vec{B}| = |\vec{C}|\), prove that : \([(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \cdot (\vec{B} + \vec{C}) = 0\).
Answer: \([(\vec{A} + \vec{B}) \times (\vec{A} + \vec{C})] \times (\vec{B} \times \vec{C}) \cdot (\vec{B} + \vec{C})\)
\(\Rightarrow [\vec{A} \vec{B} \vec{C}](C^2 - B^2) = 0 \quad (\because |\vec{C}| = |\vec{B}|)\)

Question. Consider the non zero vectors \(\vec{a}, \vec{b}, \vec{c} \, \& \, \vec{d}\) such that no three of which are coplanar then prove that \(\vec{a}[\vec{b}\vec{c}\vec{d}] + \vec{c}[\vec{a}\vec{b}\vec{d}] = \vec{b}[\vec{a}\vec{c}\vec{d}] + \vec{d}[\vec{a}\vec{b}\vec{c}]\). Hence prove that \(\vec{a}, \vec{b}, \vec{c} \, \& \, \vec{d}\) represent the position vectors of the vertices of a plane quadrilateral if \(\frac{[\vec{b}\vec{c}\vec{d}] + [\vec{a}\vec{b}\vec{d}]}{[\vec{a}\vec{c}\vec{d}] + [\vec{a}\vec{b}\vec{c}]} = 1\).
Answer: \(\vec{d} = x\vec{a} + y\vec{b} + z\vec{c}\) ....(1)
Dot with \((\vec{b} \times \vec{c}), (\vec{c} \times \vec{a}) \, \& \, (\vec{a} \times \vec{b})\) one by one:
\(x = \frac{[\vec{b} \quad \vec{c} \quad \vec{d}]}{[\vec{a} \quad \vec{b} \quad \vec{c}]}\), \(y = \frac{[\vec{c} \quad \vec{a} \quad \vec{d}]}{[\vec{a} \quad \vec{b} \quad \vec{c}]}\), \(z = \frac{[\vec{a} \quad \vec{b} \quad \vec{d}]}{[\vec{a} \quad \vec{b} \quad \vec{c}]}\)
\(\vec{d}[\vec{a} \quad \vec{b} \quad \vec{c}] + \vec{b}[\vec{a} \quad \vec{c} \quad \vec{d}] = \vec{a}[\vec{b} \quad \vec{c} \quad \vec{d}] + \vec{c}[\vec{a} \quad \vec{b} \quad \vec{d}]\)
Also, \(\vec{d} = x(\vec{a} \times \vec{b}) + y(\vec{b} \times \vec{c}) + z(\vec{c} \times \vec{a})\)
dot with \(\vec{d}\) :

Question. The base vectors \(\vec{a}_1, \vec{a}_2, \vec{a}_3\) are given in terms of base vectors \(\vec{b}_1, \vec{b}_2, \vec{b}_3\) as \(\vec{a}_1 = 2\vec{b}_1 + 3\vec{b}_2 - \vec{b}_3\) ; \(\vec{a}_2 = \vec{b}_1 - 2\vec{b}_2 + 2\vec{b}_3 \, \& \, \vec{a}_3 = 2\vec{b}_1 + \vec{b}_2 - 2\vec{b}_3\). If \(\vec{F} = 3\vec{b}_1 - \vec{b}_2 + 2\vec{b}_3\), then express \(\vec{F}\) in terms of \(\vec{a}_1, \vec{a}_2 \, \& \, \vec{a}_3\).
Answer: Let \(\vec{F} = x\vec{a}_1 + y\vec{a}_2 + z\vec{a}_3\)
\(3\vec{b}_1 - \vec{b}_2 + 2\vec{b}_3 = x\vec{a}_1 + y\vec{a}_2 + z\vec{a}_3\)
\(3 = 2x + y + 3z\) ...(i)
\(-1 = 3x - 2y - z\) ....(ii)
\(2 = -x + 2y - 2z\) ....(iii)
Solve (1), (2) & (3)
\(x = 2 \quad y = 5 \quad z = 3\)

Question. If A(\(\vec{a}\)); B(\(\vec{b}\)); C(\(\vec{c}\)) are three non collinear points, then for any point P(\(\vec{p}\)) in the plane of the \(\Delta ABC\), prove that ;
(i) \([\vec{a}\vec{b}\vec{c}] = \vec{p} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})\)
(ii) The vector \(\vec{v}\) prependicular to the plane of the triangle ABC drawn from the origin 'O' is given by \(\vec{v} = \pm \frac{[\vec{a}\vec{b}\vec{c}](\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{4\Delta^2}\) where \(\Delta\) is the vector area of the triangle ABC.
Answer: (i) \(\vec{AB} \times \vec{AC} =\)
\((\vec{a} \times \vec{b}) + (\vec{b} \times \vec{c}) + (\vec{c} \times \vec{b})\)
\(\vec{p} \cdot \vec{n} = \vec{a} \cdot \vec{n}\)
\(\vec{p} \cdot (\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}) = [\vec{a} \quad \vec{b} \quad \vec{c}]\)
(ii) Let \(\vec{v} = \lambda(\vec{AB} \times \vec{AC})\)
\(\vec{v} = \lambda(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})\)
dot with \(\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}\)
& use the relation from (1) part (1)
\([\vec{a} \quad \vec{b} \quad \vec{c}] = \lambda |2\Delta|^2\)
\(\lambda = \frac{[\vec{a} \quad \vec{b} \quad \vec{c}]}{4\Delta^2}\)
\(\vec{v} = \pm \frac{[\vec{a} \quad \vec{b} \quad \vec{c}](\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})}{4\Delta^2}\)

Question. (a) If \(\vec{p} \times \vec{x} + (\vec{x} \times \vec{a}) = \vec{b}\) ; \((\vec{p} \neq 0)\) prove that \(\vec{x} = \frac{p^2\vec{b} + (\vec{b} \cdot \vec{a})\vec{a} - p(\vec{b} \times \vec{a})}{p(p^2 + a^2)}\)
(b) Solve the following equation for the vector \(\vec{p}\) ; \(\vec{p} \times \vec{a} + (\vec{p} \cdot \vec{b})\vec{c} = \vec{b} \times \vec{c}\) where \(\vec{a}, \vec{b}, \vec{c}\) are non zero non coplanar vectors and \(\vec{a}\) is neither perpendicular to \(\vec{b}\) nor to \(\vec{c}\), hence show that \(\left(\vec{p} \times \vec{a} + \frac{[\vec{a}\vec{b}\vec{c}]}{\vec{a} \cdot \vec{c}}\vec{c}\right)\) is perpendicular to \(\vec{b} - \vec{c}\).
Answer: (a) take dot with \(\vec{a}\) \(p(\vec{x} \cdot \vec{a}) = \vec{a} \cdot \vec{b}\)
Cross with \(\vec{a}\)
\(\vec{p} \times (\vec{x} \times \vec{a}) + (\vec{x} \times \vec{a}) \times \vec{a} = \vec{b} \times \vec{a}\)
\(\vec{p}(\vec{b} - p\vec{x}) + (\vec{x} \cdot \vec{a})\vec{a} - |\vec{a}|^2 \vec{x} = \vec{b} \times \vec{a}\)
\(p\vec{b} - p^2\vec{x} + \frac{\vec{a} \cdot \vec{b}}{p}\vec{a} - |\vec{a}|^2 \vec{x} = \vec{b} \times \vec{a}\)
\(\vec{x} = \frac{p^2\vec{b} + (\vec{a} \cdot \vec{b})\vec{a} - p(\vec{b} \times \vec{a})}{p(p^2 + a^2)}\)

Question. Let the vectors \( \vec{a}, \vec{b}, \vec{c} \ \& \ \vec{d} \) be such that \( (\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \vec{0} \). Let \( P_1 \ \& \ P_2 \) be planes determined by the pairs of vectors \( \vec{a}, \vec{b} \ \& \ \vec{c}, \vec{d} \) respectively. Then the angle between \( P_1 \) and \( P_2 \) is
(a) 0
(b) \( \pi/4 \)
(c) \( \pi/3 \)
(d) \( \pi/2 \)
Answer: (a) 0

Question. If \( \vec{a}, \vec{b} \ \& \ \vec{c} \) are unit coplanar vectors, then the scalar triple product \( [2\vec{a} - \vec{b} \quad 2\vec{b} - \vec{c} \quad 2\vec{c} - \vec{a}] = \)
(a) 0
(b) 1
(c) \( -\sqrt{3} \)
(d) \( \sqrt{3} \)
Answer: (a) 0

Question. If \( \vec{a} = \hat{i} + \hat{j} - \hat{k} \), \( \vec{b} = -\hat{i} + 2\hat{j} + 2\hat{k} \ \& \ \vec{c} = -\hat{i} + 2\hat{j} - \hat{k} \), find a unit vector normal to the vectors \( \vec{a} + \vec{b} \) and \( \vec{b} - \vec{c} \). 
Answer: \( \vec{a} = (1, 1, -1); \vec{b} = (-1, 2, 2); \vec{c} = (-1, 2, -1) \)
\( \vec{a} + \vec{b} = (0, 3, 1); \quad \vec{b} - \vec{c} = (0, 0, 3) \)
Normal vector = \( \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 3 & 1 \\ 0 & 0 & 3 \end{vmatrix} = 9\hat{i} - 0\hat{j} + 0\hat{k} \)
unit normal vector = \( \pm \hat{i} \)

Question. Given that vectors \( \vec{a} \ \& \ \vec{b} \) are perpendicular to each other, find vector \( \vec{v} \) in terms of \( \vec{a} \ \& \ \vec{b} \) satisfying the equations, \( \vec{v} \cdot \vec{a} = 0 \), \( \vec{v} \cdot \vec{b} = 1 \) and \( [\vec{v} \ \vec{a} \ \vec{b}] = 1 \)
Answer: \( \vec{a} \cdot \vec{b} = 0 \) so, \( \vec{a}, \vec{b} \) and \( \vec{a} \times \vec{b} \) are perpendicular to each other.
\( \vec{V} = p\vec{a} + q\vec{b} + r(\vec{a} \times \vec{b}) \)
\( \vec{V} \cdot \vec{a} = 0 \Rightarrow p(\vec{a} \cdot \vec{a}) = 0 \Rightarrow p = 0 \)
\( \vec{V} \cdot \vec{b} = 1 \Rightarrow 0 + q|\vec{b}|^2 + 0 = 1 \Rightarrow q = \frac{1}{|\vec{b}|^2} \)
\( \vec{V} \cdot (\vec{a} \times \vec{b}) = p.0 + q.0 + r|\vec{a} \times \vec{b}|^2 \Rightarrow \frac{[\vec{a} \ \vec{b} \ \vec{v}]}{|\vec{a} \times \vec{b}|^2} = \frac{1}{|\vec{a} \times \vec{b}|^2} \Rightarrow r = \frac{1}{|\vec{a} \times \vec{b}|^2} \)
\( \vec{V} = \frac{\vec{b}}{|\vec{b}|^2} + \frac{(\vec{a} \times \vec{b})}{|\vec{a} \times \vec{b}|^2} \)

Question. \( \vec{a}, \vec{b} \ \& \ \vec{c} \) are three unit vectors such that \( \vec{a} \times (\vec{b} \times \vec{c}) = \frac{1}{2}(\vec{b} + \vec{c}) \). Find angle between vectors \( \vec{a} \ \& \ \vec{b} \) given that vectors \( \vec{b} \ \& \ \vec{c} \) are non-parallel.
Answer: \( \vec{a} \times (\vec{b} \times \vec{c}) = \vec{b} + \vec{c} \)
\( (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2}(\vec{b} + \vec{c}) \)
\( \vec{a} \cdot \vec{c} = \frac{1}{2} \quad \vec{a} \cdot \vec{b} = -\frac{1}{2} \)
\( |\vec{a}| |\vec{b}| \cos \theta = -\frac{1}{2} \Rightarrow \cos \theta = -\frac{1}{2} \Rightarrow \theta = \frac{2\pi}{3} \)

Question. The diagonals of a parallelogram are given by vectors \( 2\hat{i} + 3\hat{j} - 6\hat{k} \) and \( 3\hat{i} - 4\hat{j} - \hat{k} \). Determine its sides and also the area.
Answer: \( \vec{d}_1 = (2, 3, -6) ; \quad \vec{d}_2 = (3, -4, -1) \)
Area = \( \frac{1}{2}|\vec{d}_1 \times \vec{d}_2| = \frac{1}{2}\sqrt{1274} \)
If sides are \( \vec{a} \) and \( \vec{b} \)
\( \vec{d}_1 = \vec{a} + \vec{b} ; \quad \vec{d}_2 = \vec{a} - \vec{b} \)
\( \vec{a} = \frac{\vec{d}_1 + \vec{d}_2}{2} ; \quad \vec{b} = \frac{\vec{d}_1 - \vec{d}_2}{2} \)
\( \vec{a} = \frac{1}{2}(5, -1, -7) \quad \vec{b} = \frac{1}{2}(-1, 7, -5) \)

Question. Find the value of \( \lambda \) such that a, b, c are all non-zero and
\( (-4\hat{i} + 5\hat{j})\vec{a} + (3\hat{i} - 3\hat{j} + \hat{k})\vec{b} + (\hat{i} + \hat{j} + 3\hat{k})\vec{c} = \lambda(a\hat{i} + b\hat{j} + c\hat{k}) \)
Answer: \( (-4, 5, 0)a + (3, -3, 1)b + (1, 1, 3)c = \lambda (a, b, c) \)
\( (-4a + 3b + c - \lambda a)\hat{i} + (5a - 3b + c - \lambda b)\hat{j} + (b + 3c - \lambda c)\hat{k} = 0 \)
\( \ell\hat{i} + m\hat{j} + n\hat{k} = 0 \)
\( \ell = m = n = 0 \)
\( -4a + 3b + c - \lambda a = 0 \)
\( 5a - 3b + c - \lambda b = 0 \)
\( b + 3c - \lambda c = 0 \)
again \( (-4 - \lambda)a + 3b + c = 0 \)
\( 5a + (-3 - \lambda)b + c = 0 \)
\( 0a + 1b + (3 - \lambda)c = 0 \)
\( \begin{vmatrix} -4-\lambda & 3 & 1 \\ 5 & -3-\lambda & 1 \\ 0 & 1 & 3-\lambda \end{vmatrix} = 0 \Rightarrow \lambda = 0, -2 \pm \sqrt{29} \)

Question. Find the vector \( \vec{r} \) which is perpendicular to \( \vec{a} = \hat{i} - 2\hat{j} + 5\hat{k} \ \& \ \vec{b} = 2\hat{i} + 3\hat{j} - \hat{k} \ \& \ \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 8 = 0 \) 
Answer: \( \vec{r} = \lambda(\vec{a} \times \vec{b}) \)
\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 5 \\ 2 & 3 & -1 \end{vmatrix} \)
\( \vec{r} = \lambda(-13, -11, 7) = \hat{i}(2 - 15) - \hat{j}(-1 - 10) + \hat{k}(3 + 4) = -13\hat{i} + 11\hat{j} + 7\hat{k} \)
\( \vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) + 8 = 0 \Rightarrow \lambda = 1 \) so \( \vec{r} = (-13, -11, 7) \)

Question. If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are unit vectors, then \( |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 \) does NOT exceed [JEE 2001(Scr.), 1 + 1]
(a) 4
(b) 9
(c) 8
(d) 6
Answer: (b) 9

Question. Let \( \vec{a} = \hat{i} - \hat{k} \), \( \vec{b} = x\hat{i} + \hat{j} + (1 - x)\hat{k} \) and \( \vec{c} = y\hat{i} + x\hat{j} + (1 + x - y)\hat{k} \). Then \( [\vec{a}, \vec{b}, \vec{c}] \) depends on
(a) only x
(b) only y
(c) neither x nor y
(d) both x and y
Answer: (c) neither x nor y

Question. Let \( \vec{A}(t) = f_1(t)\hat{i} + f_2(t)\hat{j} \) and \( \vec{B}(t) = g_1(t)\hat{i} + g_2(t)\hat{j} \), \( t \in [0, 1] \), where \( f_1, f_2, g_1, g_2 \) are continuous functions. If \( \vec{A}(t) \) and \( \vec{B}(t) \) are nonzero vectors for all t and \( \vec{A}(0) = 2\hat{i} + 2\hat{j} \), \( \vec{A}(1) = 6\hat{i} + 2\hat{j} \), \( \vec{B}(0) = 3\hat{i} + 2\hat{j} \) and \( \vec{B}(1) = 2\hat{i} + 6\hat{j} \), then show that \( \vec{A}(t) \) and \( \vec{B}(t) \) are parallel for some t. 
Answer: \( \vec{A}(t) \) is parallel to \( \vec{B}(t) \) only if
\( \frac{f_1(t)}{g_1(t)} = \frac{f_2(t)}{g_2(t)} \) for some t \( \in \) [0, 1]
or \( f_1(t) \cdot g_2(t) = f_2(t) \cdot g_1(t) \)
let \( h(t) = f_1(t) \cdot g_2(t) - f_2(t) \cdot g_1(t) \)
\( h(0) = 2 \times 2 - 3 \times 3 = -5 < 0 \)
\( h(1) = f_1(1) \cdot g_2(1) - f_2(1) \cdot g_1(1) = 6 \times 6 - 2 \times 2 = 32 > 0 \)
Since, h is a continuous function and h(0).h(1)<0
\( \Rightarrow \) there is some t \( \in \) [0, 1] for which -h(t)=0

Question. If \( \vec{a} \) and \( \vec{b} \) are two unit vectors such that \( \vec{a} + 2\vec{b} \) and \( 5\vec{a} - 4\vec{b} \) are perpendicular to each other then the angle between \( \vec{a} \) and \( \vec{b} \) is 
(a) \( 45^\circ \)
(b) \( 60^\circ \)
(c) \( \cos^{-1}(1/3) \)
(d) \( \cos^{-1}(2/7) \)
Answer: (b) \( 60^\circ \)

Question. Let \( \vec{V} = 2\hat{i} + \hat{j} - \hat{k} \) and \( \vec{W} = \hat{i} + 3\hat{k} \). If \( \vec{U} \) is a unit vector, then the maximum value of the scalar triple product \( [\vec{U} \vec{V} \vec{W}] \) is
(a) -1
(b) \( \sqrt{10} + \sqrt{6} \)
(c) \( \sqrt{59} \)
(d) \( \sqrt{60} \)
Answer: (c) \( \sqrt{59} \)

Question. If \( \vec{a} = \hat{i} + a\hat{j} + \hat{k} \), \( \vec{b} = \hat{j} + a\hat{k} \), \( \vec{c} = a\hat{i} + \hat{k} \), then find the value of 'a' for which volume of parallelopiped formed by three vectors as coterminous edges, is minimum, is 
(a) \( \frac{1}{\sqrt{3}} \)
(b) \( -\frac{1}{\sqrt{3}} \)
(c) \( \pm \frac{1}{\sqrt{3}} \)
(d) None of the options
Answer: (d) None of the options

Question. If \( \vec{u}, \vec{v}, \vec{w} \) are three non-coplanar unit vectors and \( \alpha, \beta, \gamma \) are the angles between \( \vec{u} \) and \( \vec{v} \), \( \vec{v} \) and \( \vec{w} \), \( \vec{w} \) and \( \vec{u} \) respectively and \( \vec{x}, \vec{y}, \vec{z} \) are unit vectors along the bisectors of the angles \( \alpha, \beta, \gamma \) respectively. Prove that \( [\vec{x} \times \vec{y} \quad \vec{y} \times \vec{z} \quad \vec{z} \times \vec{x}] = \frac{1}{16} [\vec{u} \ \vec{v} \ \vec{w}]^2 \sec^2 \frac{\alpha}{2} \sec^2 \frac{\beta}{2} \sec^2 \frac{\gamma}{2} \). 
Answer: \( \vec{x} = \frac{\vec{u} + \vec{v}}{|\vec{u} + \vec{v}|} = \frac{1}{2} \sec \frac{\alpha}{2} (\vec{u} + \vec{v}) \)
\( \vec{y} = \frac{\vec{v} + \vec{w}}{|\vec{v} + \vec{w}|} = \frac{1}{2} \sec \frac{\beta}{2} (\vec{v} + \vec{w}) \)
\( \vec{z} = \frac{\vec{w} + \vec{u}}{|\vec{w} + \vec{u}|} = \frac{1}{2} \sec \frac{\gamma}{2} (\vec{w} + \vec{u}) \)
since \( [\vec{x} \times \vec{y} \quad \vec{y} \times \vec{z} \quad \vec{z} \times \vec{x}] = [\vec{x} \ \vec{y} \ \vec{z}]^2 = \frac{1}{64} \sec^2 \frac{\alpha}{2} \sec^2 \frac{\beta}{2} \sec^2 \frac{\gamma}{2} [\vec{u} + \vec{v} \quad \vec{v} + \vec{w} \quad \vec{w} + \vec{u}]^2 \ \dots(1) \)
\( [\vec{u} + \vec{v} \quad \vec{v} + \vec{w} \quad \vec{w} + \vec{u}]^2 = 2[\vec{u} \ \vec{v} \ \vec{w}] \)
\( [\vec{x} \times \vec{y} \quad \vec{y} \times \vec{z} \quad \vec{z} \times \vec{x}] = \frac{1}{64} \sec^2 \frac{\alpha}{2} \sec^2 \frac{\beta}{2} \sec^2 \frac{\gamma}{2} \times 4[\vec{u} \ \vec{v} \ \vec{w}]^2 = \frac{1}{16} \sec^2 \frac{\alpha}{2} \sec^2 \frac{\beta}{2} \sec^2 \frac{\gamma}{2} [\vec{u} \ \vec{v} \ \vec{w}]^2 \)

Question. A unit vector in the plane of the vectors \( 2\hat{i} + \hat{j} + \hat{k}, \hat{i} - \hat{j} + \hat{k} \) and orthogonal to \( 5\hat{i} + 2\hat{j} + 6\hat{k} \) 
(a) \( \frac{6\hat{i} - 5\hat{k}}{\sqrt{61}} \)
(b) \( \frac{3\hat{j} - \hat{k}}{\sqrt{10}} \)
(c) \( \frac{2\hat{i} - 5\hat{k}}{\sqrt{29}} \)
(d) \( \frac{2\hat{i} + \hat{j} - 2\hat{k}}{3} \)
Answer: (b) \( \frac{3\hat{j} - \hat{k}}{\sqrt{10}} \)

Question. If \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{a} \cdot \vec{b} = 1 \) and \( \vec{a} \times \vec{b} = \hat{j} - \hat{k} \), then \( \vec{b} \) equals
(a) \( \hat{i} \)
(b) \( \hat{i} - \hat{j} + \hat{k} \)
(c) \( 2\hat{j} - \hat{k} \)
(d) \( 2\hat{i} \)
Answer: (a) \( \hat{i} \)

Question. Let \( \vec{a}, \vec{b}, \vec{c}, \vec{d} \) are four distinct vectors satisfying \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \). Show that \( \vec{a} \cdot \vec{b} + \vec{c} \cdot \vec{d} \neq \vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{d} \). 
Answer: \( \vec{a} \times \vec{b} = \vec{c} \times \vec{d} \) and \( \vec{a} \times \vec{c} = \vec{b} \times \vec{d} \)
\( (\vec{a} \times \vec{b}) - (\vec{a} \times \vec{c}) \) and \( (\vec{c} \times \vec{d}) - (\vec{b} \times \vec{d}) \)
\( \vec{a} \times (\vec{b} - \vec{c}) = (\vec{c} - \vec{b}) \times \vec{d} \)
\( \vec{a} \times (\vec{b} - \vec{c}) - (\vec{c} - \vec{b}) \times \vec{d} = 0 \)
\( \vec{a} \times (\vec{b} - \vec{c}) - \vec{d} \times (\vec{b} - \vec{c}) = 0 \)
\( (\vec{a} - \vec{d}) \times (\vec{b} - \vec{c}) = 0 \)
\( \vec{a} - \vec{d} \) is || to \( (\vec{b} - \vec{c}) \)
\( (\vec{a} - \vec{d}) \cdot (\vec{b} - \vec{c}) \neq 0 \)
\( \vec{a} \cdot \vec{b} + \vec{d} \cdot \vec{c} \neq \vec{d} \cdot \vec{b} + \vec{a} \cdot \vec{c} \)

Question. Let \( \vec{a} = \hat{i} + 2\hat{j} + \hat{k}, \vec{b} = \hat{i} - \hat{j} + \hat{k} \) and \( \vec{c} = \hat{i} + \hat{j} - \hat{k} \). A vector in the plane of \( \vec{a} \) and \( \vec{b} \) whose projection on \( \vec{c} \) has the magnitude equal to \( 1/\sqrt{3} \), is 
(a) \( 4\hat{i} - \hat{j} + 4\hat{k} \)
(b) \( 3\hat{i} + \hat{j} - 3\hat{k} \)
(c) \( 2\hat{i} + \hat{j} - 2\hat{k} \)
(d) \( 4\hat{i} + \hat{j} - 4\hat{k} \)
Answer: (a) \( 4\hat{i} - \hat{j} + 4\hat{k} \)

Question. Let \( \vec{A} \) be vector parallel to line of intersection of planes \( P_1 \) and \( P_2 \) through origin. \( P_1 \) is parallel to the vectors \( 2\hat{j} + 3\hat{k} \) and \( 4\hat{j} - 3\hat{k} \) and \( P_2 \) is parallel to \( \hat{j} - \hat{k} \) and \( 3\hat{i} + 3\hat{j} \), then the angle between vector \( \vec{A} \) and \( 2\hat{i} + \hat{j} - 2\hat{k} \) is
(a) \( \pi/2 \)
(b) \( \pi/4 \)
(c) \( \pi/6 \)
(d) \( \pi/3 \)
Answer: (b) \( \pi/4 \)

Question. The number of distinct real values of \( \lambda \), for which the vectors \( -\lambda^2\hat{i} + \hat{j} + \hat{k} \), \( \hat{i} - \lambda^2\hat{j} + \hat{k} \) and \( \hat{i} + \hat{j} - \lambda^2\hat{k} \) are coplanar, is 
(a) zero
(b) one
(c) two
(d) three
Answer: (c) two

Question. Let \( \vec{a}, \vec{b}, \vec{c} \) be unit vectors such that \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \). Which one of the following is correct ?
(a) \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} = \vec{0} \)
(b) \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \neq \vec{0} \)
(c) \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{a} \times \vec{c} \neq \vec{0} \)
(d) \( \vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a} \) are mutually perpendicular.
Answer: (b) \( \vec{a} \times \vec{b} = \vec{b} \times \vec{c} = \vec{c} \times \vec{a} \neq \vec{0} \)

Question. Let the vectors \( \vec{PQ}, \vec{QR}, \vec{RS}, \vec{ST}, \vec{TU} \) and \( \vec{UP} \) represent the sides of a regular hexagon.
Statement-I : \( \vec{PQ} \times (\vec{RS} + \vec{ST}) \neq \vec{0} \)
because
Statement-II : \( \vec{PQ} \times \vec{RS} = \vec{0} \) and \( \vec{PQ} \times \vec{ST} \neq \vec{0} \)
(a) Statement-I is true, statement-II is true; statement-II is a correct explanation for statement-I
(b) Statement-I is true, statement-II is true; statement-II is NOT a correct explanation for statement-I
(c) Statement-I is true, Statement-II is False
(d) Statement-I is False, Statement-II is True
Answer: (c) Statement-I is true, Statement-II is False

Question. The edges of a parallelopiped are of unit length and are parallel to non-coplanar unit vectors \( \hat{a}, \hat{b}, \hat{c} \) such that \( \hat{a} \cdot \hat{b} = \hat{b} \cdot \hat{c} = \hat{c} \cdot \hat{a} = \frac{1}{2} \). Then the volume of the parallelopiped is 
(a) \( \frac{1}{\sqrt{2}} \)
(b) \( \frac{1}{2\sqrt{2}} \)
(c) \( \frac{\sqrt{3}}{2} \)
(d) \( \frac{1}{\sqrt{3}} \)
Answer: (a) \( \frac{1}{\sqrt{2}} \)

Question. Let two non-collinear unit vector \( \hat{a} \) and \( \hat{b} \) form an acute angle. A point P moves so that at any time t the position vector \( \vec{OP} \) (where O is the origin) is given by \( \hat{a}\cos t + \hat{b}\sin t \). When P is farthest from origin O, let M be the length of \( \vec{OP} \) and \( \hat{u} \) be the unit vector along \( \vec{OP} \). Then,
(a) \( \hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} \) and \( M = (1 + \hat{a} \cdot \hat{b})^{1/2} \)
(b) \( \hat{u} = \frac{\hat{a} - \hat{b}}{|\hat{a} - \hat{b}|} \) and \( M = (1 + \hat{a} \cdot \hat{b})^2 \)
(c) \( \hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} \) and \( M = (1 + 2\hat{a} \cdot \hat{b})^{1/2} \)
(d) \( \hat{u} = \frac{\hat{a} - \hat{b}}{|\hat{a} - \hat{b}|} \) and \( M = (1 + 2\hat{a} \cdot \hat{b})^{1/2} \)
Answer: (a) \( \hat{u} = \frac{\hat{a} + \hat{b}}{|\hat{a} + \hat{b}|} \) and \( M = (1 + \hat{a} \cdot \hat{b})^{1/2} \)

Question. If \( \vec{a}, \vec{b}, \vec{c} \) and \( \vec{d} \) are unit vectors such that \( (\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = 1 \) and \( \vec{a} \cdot \vec{c} = \frac{1}{2} \), then
(a) \( \vec{a}, \vec{b}, \vec{c} \) are non-coplanar
(b) \( \vec{b}, \vec{c}, \vec{d} \) are non-coplanar
(c) \( \vec{b}, \vec{d} \) are non-parallel
(d) \( \vec{a}, \vec{d} \) are parallel and \( \vec{b}, \vec{c} \) are parallel
Answer: (c) \( \vec{b}, \vec{d} \) are non-parallel

Question. Match the statements/expressions given in Column-I with the value given in Column-II.
Column-I
(A) Roots(s) of the equation \( 2\sin^2\theta + \sin^2 2\theta = 2 \)
(B) Points of discontinuity of the function \( f(x) = \left[ \frac{6x}{\pi} \right] \cos \left[ \frac{3x}{\pi} \right] \), where [y] denotes the largest integer less than or equal to y
(C) Volume of the parallelopiped with its edges represented by the vectors \( \hat{i} + \hat{j}, \hat{i} + 2\hat{j} \) and \( \hat{i} + \hat{j} + \pi\hat{k} \)
(D) Angle between vectors \( \vec{a} \) and \( \vec{b} \) where \( \vec{a}, \vec{b} \) and \( \vec{c} \) are unit vectors satisfying \( \vec{a} + \vec{b} + \sqrt{3}\vec{c} = \vec{0} \)
Column-II
(P) \( \pi/6 \)
(Q) \( \pi/4 \)
(R) \( \pi/3 \)
(S) \( \pi/2 \)
(T) \( \pi \)
Answer: Let \( y = \frac{3x}{\pi} \)
\( \Rightarrow \frac{1}{2} \le y \le 3 \ \forall \ x \in [\frac{\pi}{6}, \pi] \)
Now \( f(y) = [2y] \cos [y] \)
Critical points are \( y= \frac{1}{2}, y = 1, y = \frac{3}{2}, y = 3 \)
Points of discontinuity \( \left\{ \frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi \right\} \)
(C) Volume = \( \begin{vmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{vmatrix} = \pi \)
(D) \( \vec{a} + \vec{b} = -\sqrt{3}\vec{c} \Rightarrow |\vec{a} + \vec{b}|^2 = 3 \)
\( |\vec{a} + \vec{b}|^2 = 3 \Rightarrow 1 + 1 + 2 \cos \theta = 3 \)
\( \cos \theta = \frac{1}{2} \Rightarrow \theta = \frac{\pi}{3} \)

Question. If \( \vec{a} \) and \( \vec{b} \) are vectors in space given by \( \vec{a} = \frac{\hat{i} - 2\hat{j}}{\sqrt{5}} \) and \( \vec{b} = \frac{2\hat{i} + \hat{j} + 3\hat{k}}{\sqrt{14}} \), then the value of \( (2\vec{a} + \vec{b}) \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b})] \) is 
Answer: \( 2\vec{a} \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b})] + \vec{b} \cdot [(\vec{a} \times \vec{b}) \times (\vec{a} - 2\vec{b})] \)
\( = 2[\vec{a} \times (\vec{a} \times \vec{b}) \cdot (\vec{a} - 2\vec{b})] + [\vec{b} \times (\vec{a} \times \vec{b}) \cdot (\vec{a} - 2\vec{b})] \)
\( = -2\vec{b} \cdot (\vec{a} - 2\vec{b}) + (\vec{a}) \cdot (\vec{a} - 2\vec{b}) = 4|\vec{b}|^2 + |\vec{a}|^2 \)
(\( \vec{a} \cdot \vec{b} = 0 \))
= 5

Question. If \( \vec{a} \) and \( \vec{b} \) are vectors such that \( |\vec{a} + \vec{b}| = \sqrt{29} \) and \( \vec{a} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{b} \) then a possible value of \( (\vec{a} + \vec{b}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k}) \) is 
(a) 0
(b) 3
(c) 4
(d) 8
Answer: (c) 4

Question. If \( \vec{a}, \vec{b} \) and \( \vec{c} \) are unit vectors satisfying \( |\vec{a} - \vec{b}|^2 + |\vec{b} - \vec{c}|^2 + |\vec{c} - \vec{a}|^2 = 9 \), then \( |2\vec{a} + 5\vec{b} + 5\vec{c}| \) is 
Answer: Given is maximum value of expression & hence angle b/w any two is \( 120^\circ \)
\( \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} = \vec{c} \cdot \vec{a} = \frac{-1}{2} \)
so \( |2\vec{a} + 5\vec{b} + 5\vec{c}| \) is 3