RS Aggarwal Solutions for Class 12 Chapter 28 The Plane

Question 1. Find the vector equation of a plane which is at a distance of 5 units from the origin and which has \( \hat{k} \) as the unit vector normal to it.
Answer: The plane is positioned 5 units away from the origin. Since the unit normal vector is \( \hat{k} \), we use the formula \( \vec{r} \cdot \hat{n} = d \) where \( d = 5 \) and \( \hat{n} = \hat{k} \). The vector equation becomes \( \vec{r} \cdot \hat{k} = 5 \). When we express \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), we get \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot \hat{k} = 5 \), which simplifies to \( z = 5 \). This is the Cartesian form of the plane.

Exam Tip: When the unit normal vector is given directly, use it immediately in the formula \( \vec{r} \cdot \hat{n} = d \) without further normalization. The Cartesian equation will match the vector equation when you substitute and simplify.

Question 2. Find the vector and Cartesian equations of a plane which is at a distance of 7 units from the origin and whose normal vector from the origin is \( (3\hat{i} + 5\hat{j} - 6\hat{k}) \).
Answer: The distance from the origin is \( d = 7 \) and the normal vector is \( \vec{n} = 3\hat{i} + 5\hat{j} - 6\hat{k} \). First, we determine the unit normal vector by dividing by the magnitude of \( \vec{n} \). The magnitude is \( |\vec{n}| = \sqrt{3^2 + 5^2 + (-6)^2} = \sqrt{9 + 25 + 36} = \sqrt{70} \). Therefore, the unit normal is \( \hat{n} = \frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}} \). Using \( \vec{r} \cdot \hat{n} = d \), the vector equation is \( \vec{r} \cdot \left(\frac{3\hat{i} + 5\hat{j} - 6\hat{k}}{\sqrt{70}}\right) = 7 \), which simplifies to \( \vec{r} \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 7\sqrt{70} \). For the Cartesian form, let \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \). Then \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (3\hat{i} + 5\hat{j} - 6\hat{k}) = 3x + 5y - 6z \), giving us the Cartesian equation \( 3x + 5y - 6z = 7\sqrt{70} \).

Exam Tip: Always normalize the given normal vector to a unit vector before applying the plane equation formula. This ensures your vector and Cartesian forms are consistent with the given distance.

Question 3. Find the vector and Cartesian equations of a plane which is at a distance of \( \frac{6}{\sqrt{29}} \) from the origin and whose normal vector from the origin is \( (2\hat{i} - 3\hat{j} + 4\hat{k}) \).
Answer: We have \( d = \frac{6}{\sqrt{29}} \) and \( \vec{n} = 2\hat{i} - 3\hat{j} + 4\hat{k} \). The magnitude of the normal vector is \( |\vec{n}| = \sqrt{2^2 + (-3)^2 + 4^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \). The unit normal vector is \( \hat{n} = \frac{2\hat{i} - 3\hat{j} + 4\hat{k}}{\sqrt{29}} \). Using the plane equation \( \vec{r} \cdot \hat{n} = d \), we get \( \vec{r} \cdot \left(\frac{2\hat{i} - 3\hat{j} + 4\hat{k}}{\sqrt{29}}\right) = \frac{6}{\sqrt{29}} \). Multiplying both sides by \( \sqrt{29} \), the vector equation becomes \( \vec{r} \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 6 \). For the Cartesian equation, substituting \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \) gives us \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - 3\hat{j} + 4\hat{k}) = 2x - 3y + 4z \), so the Cartesian form is \( 2x - 3y + 4z = 6 \).

Exam Tip: When the distance is already expressed as a fraction with a square root, verify that after normalization and simplification the final equations have integer or simple coefficients.

Question 4. Find the vector and Cartesian equations of a plane which is at a distance of 6 units from the origin and which has a normal with direction ratios 2, -1, -2.
Answer: The distance is \( d = 6 \) and the direction ratios of the normal are 2, -1, -2. We express the normal vector as \( \vec{n} = 2\hat{i} - \hat{j} - 2\hat{k} \). The magnitude is \( |\vec{n}| = \sqrt{2^2 + (-1)^2 + (-2)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \). The unit normal vector is \( \hat{n} = \frac{2\hat{i} - \hat{j} - 2\hat{k}}{3} \). Using the formula \( \vec{r} \cdot \hat{n} = d \), we get \( \vec{r} \cdot \left(\frac{2\hat{i} - \hat{j} - 2\hat{k}}{3}\right) = 6 \). Multiplying both sides by 3, the vector equation is \( \vec{r} \cdot (2\hat{i} - \hat{j} - 2\hat{k}) = 18 \). For the Cartesian form, with \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), we compute \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (2\hat{i} - \hat{j} - 2\hat{k}) = 2x - y - 2z \), giving the Cartesian equation \( 2x - y - 2z = 18 \).

Exam Tip: When direction ratios are given instead of a vector, convert them to a vector form and then calculate the magnitude to normalize. This two-step process ensures accuracy.

Question 5. Find the vector and Cartesian equations of a plane which passes through the point (1, 4, 6) and normal vector to the plane is \( (\hat{i} - 2\hat{j} + \hat{k}) \).
Answer: The plane passes through point \( A = (1, 4, 6) \), so the position vector is \( \vec{a} = \hat{i} + 4\hat{j} + 6\hat{k} \). The normal vector is \( \vec{n} = \hat{i} - 2\hat{j} + \hat{k} \). The vector equation of a plane passing through a point with a given normal is \( (\vec{r} - \vec{a}) \cdot \vec{n} = 0 \), which can be rewritten as \( \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n} \). We calculate \( \vec{a} \cdot \vec{n} = (\hat{i} + 4\hat{j} + 6\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = (1)(1) + (4)(-2) + (6)(1) = 1 - 8 + 6 = -1 \). Therefore, the vector equation is \( \vec{r} \cdot (\hat{i} - 2\hat{j} + \hat{k}) = -1 \). For the Cartesian form, with \( \vec{r} = x\hat{i} + y\hat{j} + z\hat{k} \), we have \( (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} - 2\hat{j} + \hat{k}) = x - 2y + z \), giving the Cartesian equation \( x - 2y + z = -1 \).

Exam Tip: Always verify your answer by substituting the given point back into the Cartesian equation to confirm it satisfies the plane. For this problem, substituting (1, 4, 6) should yield -1 on the right-hand side.