RS Aggarwal Solutions for Class 12 Chapter 24 Cross, or Vector, Product of Vectors

Question 1. Find \( \vec{a} \times \vec{b} \) and \( |\vec{a} \times \vec{b}| \), when
(i) \( \vec{a} = \hat{i} - \hat{j} + 2\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} - 4\hat{k} \)
(ii) \( \vec{a} = 2\hat{i} - \hat{j} + 3\hat{k} \) and \( \vec{b} = 3\hat{i} + 5\hat{j} - 2\hat{k} \)
(iii) \( \vec{a} = \hat{i} - 7\hat{j} + 7\hat{k} \) and \( \vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k} \)
(iv) \( \vec{a} = 4\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{b} = 3\hat{i} + \hat{k} \)
(v) \( \vec{a} = 3\hat{i} + 4\hat{j} \) and \( \vec{b} = \hat{i} + \hat{j} + \hat{k} \)
Answer: Using the formula \( \vec{a} \times \vec{b} = (a_2b_3 - b_2a_3)\hat{i} + (a_3b_1 - b_3a_1)\hat{j} + (a_1b_2 - b_1a_2)\hat{k} \):

(i) Here, \( a_1 = 1, a_2 = -1, a_3 = 2 \) and \( b_1 = 2, b_2 = 3, b_3 = -4 \). Substituting these values:
\( \vec{a} \times \vec{b} = ((-1) \times (-4) - 3 \times 2)\hat{i} + (2 \times 2 - (-4) \times 1)\hat{j} + (1 \times 3 - 2 \times (-1))\hat{k} = (-2\hat{i} + 8\hat{j} + 5\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(-2)^2 + 8^2 + 5^2} = \sqrt{93} \)

(ii) Here, \( a_1 = 2, a_2 = -1, a_3 = 3 \) and \( b_1 = 3, b_2 = 5, b_3 = -2 \). Substituting these values:
\( \vec{a} \times \vec{b} = ((-1) \times (-2) - 5 \times 3)\hat{i} + (3 \times 3 - (-2) \times 2)\hat{j} + (2 \times 5 - 3 \times (-1))\hat{k} = (-17\hat{i} + 13\hat{j} + 7\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(-17)^2 + 13^2 + 7^2} = 13\sqrt{3} \)

(iii) Here, \( a_1 = 1, a_2 = -7, a_3 = 7 \) and \( b_1 = 3, b_2 = -2, b_3 = 2 \). Substituting these values:
\( \vec{a} \times \vec{b} = ((-7) \times 2 - (-2) \times 7)\hat{i} + (7 \times 3 - 1 \times 2)\hat{j} + ((-2) \times 1 - 3 \times (-7))\hat{k} = (0\hat{i} + 19\hat{j} + 19\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = 19\sqrt{2} \)

(iv) Here, \( a_1 = 4, a_2 = 1, a_3 = -2 \) and \( b_1 = 3, b_2 = 0, b_3 = 1 \). Substituting these values:
\( \vec{a} \times \vec{b} = (1 \times 1 - (0) \times (-2))\hat{i} + ((-2) \times 3 - 1 \times 4)\hat{j} + (4 \times 0 - 3 \times 1)\hat{k} = (\hat{i} - 10\hat{j} - 3\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + (-10)^2 + (-3)^2} = \sqrt{110} \)

(v) Here, \( a_1 = 3, a_2 = 4, a_3 = 0 \) and \( b_1 = 1, b_2 = 1, b_3 = 1 \). Substituting these values:
\( \vec{a} \times \vec{b} = (4 \times 1 - 1 \times 0)\hat{i} + (0 \times 1 - 1 \times 3)\hat{j} + (3 \times 1 - 1 \times 4)\hat{k} = (4\hat{i} - 3\hat{j} - \hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{4^2 + (-3)^2 + (-1)^2} = \sqrt{26} \)
In simple words: The cross product gives you a new vector that points perpendicular to both original vectors. Its length tells you how "perpendicular" they are - if they're parallel, the length is zero.

Exam Tip: Always arrange components in order (i, j, k) before applying the formula. Double-check signs when substituting negative values.

Question 2. Find \( \lambda \) if \( (2\hat{i} + 6\hat{j} + 14\hat{k}) \times (\hat{i} - \lambda\hat{j} + 7\hat{k}) = \vec{0} \).
Answer: For two vectors to have a zero cross product, they must be parallel (one is a scalar multiple of the other). Setting up the condition for parallelism, the cross product formula gives us three component equations. Working through the substitution with \( a_1 = 2, a_2 = 6, a_3 = 14 \) and \( b_1 = 1, b_2 = -\lambda, b_3 = 7 \):
\( \vec{a} \times \vec{b} = (6 \times 7 - (-\lambda) \times 14)\hat{i} + (14 \times 1 - 2 \times 7)\hat{j} + (2 \times (-\lambda) - 1 \times 6)\hat{k} = \vec{0} \)
From the j-component: \( 14 - 14 = 0 \) ✓
From the i-component: \( 42 + 14\lambda = 0 \)
\( \implies \lambda = -3 \)
In simple words: Two vectors are parallel when their cross product equals zero. This happens when one vector is just a scaled version of the other.

Exam Tip: For parallel vectors, corresponding components stay proportional - check this ratio to find unknown scalars quickly.

Question 3. If \( \vec{a} = (-3\hat{i} + 4\hat{j} - 7\hat{k}) \) and \( \vec{b} = (6\hat{i} + 2\hat{j} - 3\hat{k}) \), find \( (\vec{a} \times \vec{b}) \). Verify that (i) \( \vec{a} \) and \( (\vec{a} \times \vec{b}) \) are perpendicular to each other and (ii) \( \vec{b} \) and \( (\vec{a} \times \vec{b}) \) are perpendicular to each other.
Answer: We compute the cross product: with \( a_1 = -3, a_2 = 4, a_3 = -7 \) and \( b_1 = 6, b_2 = 2, b_3 = -3 \), we get
\( \vec{a} \times \vec{b} = (4 \times (-3) - 2 \times (-7))\hat{i} + ((-7) \times 6 - (-3) \times (-3))\hat{j} + ((-3) \times 2 - 6 \times 4)\hat{k} = (2\hat{i} - 51\hat{j} - 30\hat{k}) \)

For (i): To verify that \( \vec{a} \) and \( (\vec{a} \times \vec{b}) \) are perpendicular, we check their dot product:
\( \vec{a} \cdot (\vec{a} \times \vec{b}) = (-3)(2) + (4)(-51) + (-7)(-30) = -6 - 204 + 210 = 0 \) ✓

For (ii): Similarly, for \( \vec{b} \) and \( (\vec{a} \times \vec{b}) \):
\( \vec{b} \cdot (\vec{a} \times \vec{b}) = (6)(2) + (2)(-51) + (-3)(-30) = 12 - 102 + 90 = 0 \) ✓

Since both dot products equal zero, both vectors are perpendicular to their cross product, which is a fundamental property of the cross product operation.In simple words: A cross product always creates a vector that stands at right angles to both vectors you started with - this is what makes it so useful in geometry and physics.

Exam Tip: Always verify perpendicularity by computing dot products - they must equal zero. This confirms your cross product calculation is correct.

Question 4. Find the value of:
(i) \( (\hat{i} \times \hat{j}) \cdot \hat{k} + \hat{i} \cdot \hat{j} \)
(ii) \( (\hat{j} \times \hat{k}) \cdot \hat{i} + \hat{j} \cdot \hat{k} \)
(iii) \( \hat{i} \times (\hat{j} + \hat{k}) + \hat{j} \times (\hat{k} + \hat{i}) + \hat{k} \times (\hat{i} + \hat{j}) \)
Answer:
(i) We know that \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{i} \cdot \hat{j} = 0 \). Therefore:
\( (\hat{i} \times \hat{j}) \cdot \hat{k} + \hat{i} \cdot \hat{j} = \hat{k} \cdot \hat{k} + 0 = 1 \)

(ii) We know that \( \hat{j} \times \hat{k} = \hat{i} \) and \( \hat{j} \cdot \hat{k} = 0 \). Therefore:
\( (\hat{j} \times \hat{k}) \cdot \hat{i} + \hat{j} \cdot \hat{k} = \hat{i} \cdot \hat{i} + 0 = 1 \)

(iii) Using the given properties \( \hat{i} \times \hat{k} = -\hat{j} \), \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{j} \times \hat{i} = -\hat{k} \), \( \hat{k} \times \hat{i} = \hat{j} \), and \( \hat{k} \times \hat{j} = -\hat{i} \), we expand:
\( \hat{i} \times (\hat{j} + \hat{k}) + \hat{j} \times (\hat{k} + \hat{i}) + \hat{k} \times (\hat{i} + \hat{j}) = (\hat{k} - \hat{j}) + (\hat{i} - \hat{k}) + (\hat{j} - \hat{i}) = 0 \)
In simple words: Unit vector cross products follow a fixed pattern - if you go around the cycle i, j, k, you get the next one, and going backward gives you the negative.

Exam Tip: Memorize the cyclic property of unit vectors: \( \hat{i} \times \hat{j} = \hat{k} \), \( \hat{j} \times \hat{k} = \hat{i} \), \( \hat{k} \times \hat{i} = \hat{j} \). Reversing the order flips the sign.

Question 5. Find the unit vectors perpendicular to both \( \vec{a} \) and \( \vec{b} \) when
(i) \( \vec{a} = 3\hat{i} + \hat{j} - 2\hat{k} \) and \( \vec{b} = 2\hat{i} + 3\hat{j} - \hat{k} \)
(ii) \( \vec{a} = \hat{i} - 2\hat{j} + 3\hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - \hat{k} \)
(iii) \( \vec{a} = \hat{i} + 3\hat{j} - 2\hat{k} \) and \( \vec{b} = -\hat{i} + 3\hat{k} \)
(iv) \( \vec{a} = 4\hat{i} + 2\hat{j} - \hat{k} \) and \( \vec{b} = \hat{i} + 4\hat{j} - \hat{k} \)
Answer: A vector perpendicular to both given vectors is obtained from their cross product. The unit vector is found by dividing the cross product by its magnitude and taking both positive and negative directions.

(i) With \( a_1 = 3, a_2 = 1, a_3 = -2 \) and \( b_1 = 2, b_2 = 3, b_3 = -1 \):
\( \vec{a} \times \vec{b} = (1 \times (-1) - 3 \times (-2))\hat{i} + ((-2) \times 2 - (-1) \times 3)\hat{j} + (3 \times 3 - 2 \times 1)\hat{k} = (5\hat{i} - \hat{j} + 7\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(5)^2 + (-1)^2 + (7)^2} = 5\sqrt{3} \)
Unit vector: \( \vec{r} = \pm \frac{5\hat{i} - \hat{j} + 7\hat{k}}{5\sqrt{3}} \)

(ii) With \( a_1 = 1, a_2 = -2, a_3 = 3 \) and \( b_1 = 1, b_2 = 2, b_3 = -1 \):
\( \vec{a} \times \vec{b} = ((-2) \times (-1) - 2 \times 3)\hat{i} + (3 \times 1 - (-1) \times 1)\hat{j} + (1 \times 2 - (-2) \times 1)\hat{k} = (-4\hat{i} + 4\hat{j} + 4\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(-4)^2 + 4^2 + 4^2} = 4\sqrt{3} \)
Unit vector: \( \vec{r} = \pm \frac{-4\hat{i} + 4\hat{j} + 4\hat{k}}{4\sqrt{3}} \)

(iii) With \( a_1 = 1, a_2 = 3, a_3 = -2 \) and \( b_1 = -1, b_2 = 0, b_3 = 3 \):
\( \vec{a} \times \vec{b} = (3 \times 3 - 0 \times (-2))\hat{i} + ((-2) \times (-1) - 3 \times 1)\hat{j} + (1 \times 0 - (-1) \times 3)\hat{k} = (9\hat{i} - \hat{j} + 3\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(9)^2 + (-1)^2 + (3)^2} = \sqrt{91} \)
Unit vector: \( \vec{r} = \pm \frac{9\hat{i} - \hat{j} + 3\hat{k}}{\sqrt{91}} \)

(iv) With \( a_1 = 4, a_2 = 2, a_3 = -1 \) and \( b_1 = 1, b_2 = 4, b_3 = -1 \):
\( \vec{a} \times \vec{b} = (2 \times (-1) - 4 \times (-1))\hat{i} + ((-1) \times 1 - (-1) \times 4)\hat{j} + (4 \times 4 - 1 \times 2)\hat{k} = (-\hat{i} + \hat{j} + 4\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(-1)^2 + 1^2 + 4^2} = 4\sqrt{3} \)
Unit vector: \( \vec{r} = \pm \frac{-\hat{i} + \hat{j} + 4\hat{k}}{4\sqrt{3}} \)
In simple words: The cross product gives you a direction perpendicular to both vectors. Normalize it by dividing by its length to get a unit vector pointing in that direction.

Exam Tip: Always compute the magnitude after finding the cross product. The ± sign indicates there are two perpendicular directions - choose based on the right-hand rule or problem context.

Question 6. Find the unit vectors perpendicular to the plane of the vectors \( \vec{a} = 2\hat{i} - 6\hat{j} - 3\hat{k} \) and \( \vec{b} = 4\hat{i} + 3\hat{j} - \hat{k} \).
Answer: A vector perpendicular to the plane containing both vectors is their cross product. With \( a_1 = 2, a_2 = -6, a_3 = -3 \) and \( b_1 = 4, b_2 = 3, b_3 = -1 \):
\( \vec{a} \times \vec{b} = ((-6) \times (-1) - 3 \times (-3))\hat{i} + ((-3) \times 4 - (-1) \times 2)\hat{j} + (2 \times 3 - 4 \times (-6))\hat{k} = (2\hat{i} + 3\hat{j} + 14\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{(2)^2 + (3)^2 + (14)^2} = \sqrt{209} \)
The unit vectors perpendicular to the plane are:
\( \vec{r} = \pm \frac{2\hat{i} + 3\hat{j} + 14\hat{k}}{\sqrt{209}} \)
In simple words: Any plane has exactly two directions perpendicular to it - one pointing up and one pointing down. We find both by computing the cross product and normalizing.

Exam Tip: The phrase "perpendicular to the plane" always means use the cross product of the two vectors in that plane. The ± symbol shows both normal directions exist.

Question 8. Find a vector of magnitude 5 units, perpendicular to each of the vectors \( (\vec{a} + \vec{b}) \) and \( (\vec{a} - \vec{b}) \), where \( \vec{a} = (\hat{i} + \hat{j} + \hat{k}) \) and \( \vec{b} = (\hat{i} + 2\hat{j} + 3\hat{k}) \).
Answer: First, compute the required vector sums:
\( \vec{a} + \vec{b} = (1 + 1)\hat{i} + (1 + 2)\hat{j} + (1 + 3)\hat{k} = 2\hat{i} + 3\hat{j} + 4\hat{k} \)
\( \vec{a} - \vec{b} = (1 - 1)\hat{i} + (1 - 2)\hat{j} + (1 - 3)\hat{k} = 0\hat{i} - \hat{j} - 2\hat{k} \)

Next, find the cross product \( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) \). With components \( (a + b) = (2, 3, 4) \) and \( (a - b) = (0, -1, -2) \):
\( (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = (3 \times (-2) - (-1) \times 4)\hat{i} + (4 \times 0 - 2 \times (-2))\hat{j} + (2 \times (-1) - 0 \times 3)\hat{k} = (-2\hat{i} + 4\hat{j} - 2\hat{k}) \)
\( |(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})| = \sqrt{(-2)^2 + 4^2 + (-2)^2} = \sqrt{24} = 2\sqrt{6} \)

The unit vector is \( \pm \frac{-2\hat{i} + 4\hat{j} - 2\hat{k}}{2\sqrt{6}} = \pm \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} \)

For magnitude 5, scale by 5:
\( \vec{r} = \pm 5 \cdot \frac{-\hat{i} + 2\hat{j} - \hat{k}}{\sqrt{6}} = \pm \frac{-5\hat{i} + 10\hat{j} - 5\hat{k}}{\sqrt{6}} \)
In simple words: When you need a vector perpendicular to two combined vectors, treat the sum and difference as your new pair, compute their cross product, then scale to the target magnitude.

Exam Tip: Double-check your vector sums before crossing - a small addition error early will ruin your final answer. Simplify the cross product components before calculating magnitude.

Question 9. Find the angle between two vectors \( \vec{a} \) and \( \vec{b} \) with magnitudes 1 and 2 respectively and \( |\vec{a} \times \vec{b}| = \sqrt{3} \).
Answer: Using the formula for the magnitude of a cross product:
\( |\vec{a} \times \vec{b}| = |\vec{a}| \cdot |\vec{b}| \sin\theta = \sqrt{3} \)

Substitute the given magnitudes:
\( 1 \times 2 \times \sin\theta = \sqrt{3} \)
\( 2\sin\theta = \sqrt{3} \)
\( \sin\theta = \frac{\sqrt{3}}{2} \)

Taking the inverse sine:
\( \theta = \sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3} \text{ (or } 60°\text{)} \)
In simple words: The cross product's size tells you the angle between vectors. If the cross product is large, they're nearly perpendicular; if small, they're nearly parallel.

Exam Tip: Memorize the special angle values: \( \sin(60°) = \frac{\sqrt{3}}{2} \), \( \sin(45°) = \frac{1}{\sqrt{2}} \), \( \sin(30°) = \frac{1}{2} \). They appear frequently in cross product problems.

Question 10. If \( \vec{a} = (\hat{i} - \hat{j}) \), \( \vec{b} = (3\hat{j} - \hat{k}) \) and \( \vec{c} = (7\hat{i} - \hat{k}) \), find a vector \( \vec{d} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and for which \( \vec{c} \cdot \vec{d} = 1 \).
Answer: Any vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) can be written as \( \vec{d} = k.(\vec{a} \times \vec{b}) \), where k is a scalar. With \( a_1 = 1, a_2 = -1, a_3 = 0 \) and \( b_1 = 0, b_2 = 3, b_3 = -1 \):
\( \vec{a} \times \vec{b} = ((-1) \times (-1) - 3 \times 0)\hat{i} + (0 \times 0 - (-1) \times 1)\hat{j} + (1 \times 3 - 0 \times (-1))\hat{k} = (\hat{i} + \hat{j} + 3\hat{k}) \)
\( |\vec{a} \times \vec{b}| = \sqrt{1^2 + 1^2 + 3^2} = \sqrt{11} \)

So \( \vec{d} = k.(\hat{i} + \hat{j} + 3\hat{k}) \) for some scalar k. Using the condition \( \vec{c} \cdot \vec{d} = 1 \):
\( (7\hat{i} - \hat{k}) \cdot [k(\hat{i} + \hat{j} + 3\hat{k})] = 1 \)
\( k[(7)(1) + (0)(1) + (-1)(3)] = 1 \)
\( k(7 - 3) = 1 \)
\( 4k = 1 \)
\( k = \frac{1}{4} \)

Therefore:
\( \vec{d} = \frac{1}{4}(\hat{i} + \hat{j} + 3\hat{k}) = \frac{\hat{i} + \hat{j} + 3\hat{k}}{4} \)
In simple words: Build the perpendicular direction from the cross product, then use the extra constraint (the dot product condition) to pin down exactly which scaled version you need.

Exam Tip: When a vector has a perpendicularity condition AND another constraint like a dot product or magnitude, solve in two stages: first get the perpendicular direction, then use the second condition to find the scaling factor.

Question 11. If \( \vec{a} = (4\hat{i} + 5\hat{j} - \hat{k}) \), \( \vec{b} = (\hat{i} - 4\hat{j} + 5\hat{k}) \) and \( \vec{c} = (3\hat{i} + \hat{j} - \hat{k}) \), find a vector \( \vec{d} \) which is perpendicular to both \( \vec{a} \) and \( \vec{b} \) and for which \( \vec{c} \cdot \vec{d} = 21 \).
Answer: Express any vector perpendicular to both \( \vec{a} \) and \( \vec{b} \) as \( \vec{d} = k.(\vec{a} \times \vec{b}) \), where k is a scalar. With \( a_1 = 4, a_2 = 5, a_3 = -1 \) and \( b_1 = 1, b_2 = -4, b_3 = 5 \):
\( \vec{a} \times \vec{b} = (5 \times 5 - (-4) \times (-1))\hat{i} + ((-1) \times 1 - 5 \times 4)\hat{j} + (4 \times (-4) - 1 \times 5)\hat{k} = (25 - 4)\hat{i} + (-1 - 20)\hat{j} + (-16 - 5)\hat{k} = (21\hat{i} - 21\hat{j} - 21\hat{k}) \)
\( = 21(\hat{i} - \hat{j} - \hat{k}) \)

So \( \vec{d} = k(\hat{i} - \hat{j} - \hat{k}) \) for some scalar k. Using \( \vec{c} \cdot \vec{d} = 21 \):
\( (3\hat{i} + \hat{j} - \hat{k}) \cdot [k(\hat{i} - \hat{j} - \hat{k})] = 21 \)
\( k[(3)(1) + (1)(-1) + (-1)(-1)] = 21 \)
\( k[3 - 1 + 1] = 21 \)
\( 3k = 21 \)
\( k = 7 \)

Therefore:
\( \vec{d} = 7(\hat{i} - \hat{j} - \hat{k}) = (7\hat{i} - 7\hat{j} - 7\hat{k}) \)
In simple words: The cross product creates a direction perpendicular to both starting vectors. The dot product condition then tells you the exact multiple of that direction you need to match the required value.

Exam Tip: Factor out any common multiples from the cross product components - this simplifies the dot product calculation considerably. Always verify your answer by checking both the perpendicularity (zero dot products with a and b) and the given constraint.

Question 12. Prove that \( |\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta \), where θ is the angle between \( \vec{a} \) and \( \vec{b} \).
Answer: We begin by recalling that the magnitude of the dot product is given by \( |\vec{a} \cdot \vec{b}| = |\vec{a}||\vec{b}|\cos\theta \). Similarly, the magnitude of the cross product satisfies \( |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta \). We also know that \( \tan\theta = \frac{|\vec{a} \times \vec{b}|}{|\vec{a} \cdot \vec{b}|} \). Multiplying both sides by \( |\vec{a} \cdot \vec{b}| \) gives us \( |\vec{a} \times \vec{b}| = (\vec{a} \cdot \vec{b}) \tan \theta \). This completes the proof.
In simple words: The cross product magnitude relates to the dot product through the tangent of the angle between them. You can find the cross product size by multiplying the dot product by the tangent value.

Exam Tip: Always recall the standard formulas for dot and cross product magnitudes before manipulating them - this direct substitution approach avoids lengthy recalculation.

Question 13. Write the value of p for which \( \vec{a} = (3\vec{i} + 2\vec{j} + 9\vec{k}) \) and \( \vec{b} = (\vec{i} + p\vec{j} + 3\vec{k}) \) are parallel vectors.
Answer: For two vectors to be parallel, their cross product must equal zero: \( \vec{a} \times \vec{b} = 0 \). Using the given vectors where \( a_1 = 3, a_2 = 2, a_3 = 9 \) and \( b_1 = 1, b_2 = p, b_3 = 3 \), we apply the cross product formula: \( \vec{a} \times \vec{b} = (a_2b_3 - b_2a_3)\vec{i} + (a_3b_1 - b_3a_1)\vec{j} + (a_1b_2 - b_1a_2)\vec{k} \). Substituting these values yields: \( \vec{a} \times \vec{b} = (6 - 9p)\vec{i} + (0)\vec{j} + (3p - 2)\vec{k} = 0 \). For the cross product to be zero, the coefficient of \( \vec{i} \) must vanish: \( 6 - 9p = 0 \). Solving for p gives us \( p = \frac{2}{3} \).
In simple words: Two vectors point in the same direction when their cross product becomes zero. Set the first component equation to zero and solve to find that p equals two-thirds.

Exam Tip: For parallel vectors, setting at least one component of the cross product to zero is sufficient - you don't need all components to vanish if one alone determines the unknown parameter.