Class 12 Maths MCQ – Three Dimensional Geometry – Plane

This set of Class 12 Maths Chapter 11 Multiple Choice Questions & Answers (MCQs) focuses on “Three Dimensional Geometry – Plane”.

1. Which of the following is not the correct formula for representing a plane?
a) $$\vec{r}.\hat{n}=d$$
b) ax+by+cz=d
c) lx+my+nz=d
d) al+mb+cn=d2

Explanation: $$\vec{r}.\hat{n}=d$$, ax+by+cz=d, lx+my+nz=d are the various ways of representing a plane.
$$\vec{r}.\hat{n}$$=d is the vector equation of the plane, where $$\hat{n}$$ is the unit vector normal to the plane.
ax+by+cz=d, lx+my+nz=d are the Cartesian equation of the plane in the normal form where, a, b, c are the direction ratios and l, m, n are the direction cosines of the normal to the plane respectively.

2. If the plane passes through three collinear points $$(x_1,y_1,z_1),(x_2,y_2,z_2),(x_3,y_3,z_3)$$ then which of the following is true?
a) $$x_1 y_1 z_1+x_2 y_2 z_2+x_3 y_3 z_3$$=0
b) $$\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}$$=0
c) $$\begin{vmatrix}x_1\\y_2\\z_3\end{vmatrix}$$=0
d) $$x_1 x_2 x_3+y_1 y_2 y_3+z_1 z_2 z_3=0$$

Explanation: If the three points are collinear, then they will be in a straight line and hence the determinant of the three points will be zero.
i.e.$$\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}$$=0

3. Find the vector equation of the plane which is at a distance of $$\frac{7}{\sqrt{38}}$$ from the origin and the normal vector from origin is $$2\hat{i}+3\hat{j}-5\hat{k}$$?
a) $$\vec{r}.(\frac{2\hat{i}}{38}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{56}}$$
b) $$\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}$$
c) $$\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{5\hat{j}}{\sqrt{38}}+\frac{3\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}$$
d) $$\vec{r}.(\frac{2\hat{i}}{\sqrt{58}}-\frac{3\hat{j}}{\sqrt{37}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}$$

Explanation: Let $$\vec{n}=2\hat{i}+3\hat{j}-5\hat{k}$$
$$\hat{n}=\frac{\vec{n}}{|\vec{n}|}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{(2^2+3^2+(-5)^2)}}=\frac{(2\hat{i}+3\hat{j}-5\hat{k})}{\sqrt{38}}$$
Hence, the required equation of the plane is $$\vec{r}.(\frac{2\hat{i}}{\sqrt{38}}+\frac{3\hat{j}}{\sqrt{38}}-\frac{5\hat{k}}{\sqrt{38}})=\frac{7}{\sqrt{38}}$$

4. Find the equation of the plane passing through three points (1,2,-1), (0,-1,2) and (3,1,1).
a) $$(\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(-\hat{j}+2\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]$$=0
b) $$(\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]$$=0
c) $$(\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(3\hat{i}+\hat{j}+\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]$$=0
d) $$(\vec{r}-(\hat{i}+2\hat{j})).[(-\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]$$=0

Explanation: Let $$\vec{a}=\hat{i}+2\hat{j}-\hat{k}, \,\vec{b}=-\hat{j}+2\hat{k}, \,\vec{c}=3\hat{i}+\hat{j}+\hat{k}$$
The vector equation of the plane passing through three points is given by
$$(\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]$$=0
$$(\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[((-\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-\hat{k}))×((3\hat{i}+\hat{j}+\hat{k})–(\hat{i}+2\hat{j}-\hat{k}))]$$=0
$$(\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]$$=0

5. Find the Cartesian equation of the plane $$\vec{r}.(2\hat{i}+\hat{j}-\hat{k})$$=4.
a) x+y-z=-4
b) 2x+y-z=4
c) x+y+z=4
d) -2x-y+z=4

Explanation: Given that the equation of the plane is $$\vec{r}.(2\hat{i}+\hat{j}-\hat{k})$$=4
We know that, $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
∴$$(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}+\hat{j}-\hat{k})$$=4
⇒2x+y-z=4 is the Cartesian equation of the plane.

6. Find the vector equation of the plane passing through the point (2,1,-1) and normal to the plane is $$2\hat{i}+\hat{j}-3\hat{k}$$?
a) $$(\vec{r}-(2\hat{i}-7\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})$$=0
b) $$(\vec{r}-(2\hat{i}+3\hat{j}-\hat{k})).(2\hat{i}-3\hat{k})$$=0
c) $$(\vec{r}-(\hat{i}+\hat{j}-3\hat{k})).(2\hat{i}+6\hat{j}-3\hat{k})$$=0
d) $$(\vec{r}-(2\hat{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})$$=0

Explanation: The position vector of the point (2,1,-1) is $$\vec{a}=2\hat{i}+\hat{j}-\hat{k}$$ and the normal vector $$\vec{N}$$ perpendicular to the plane is $$\vec{N}=2\hat{i}+\hat{j}-3\hat{k}$$
The vector equation of the plane is given by $$(\vec{r}-\vec{a}).\vec{N}$$=0
Therefore, $$(\vec{r}-(2\hat{i}+\hat{j}-\hat{k})).(2\hat{i}+\hat{j}-3\hat{k})$$=0

7. Find the distance of the plane 3x+4y-5z-7=0.
a) $$\frac{7}{\sqrt{40}}$$
b) $$\frac{6}{\sqrt{34}}$$
c) $$\frac{8}{\sqrt{50}}$$
d) $$\frac{7}{\sqrt{50}}$$

Explanation: From the given equation, the direction ratios of the normal to the plane are 3, 4, -5; the direction cosines are
$$\frac{3}{\sqrt{3^2+4^2+(-5)^2}},\frac{4}{\sqrt{3^2+4^2+(-5)^2}},\frac{-5}{\sqrt{3^2+4^2+(-5)^2}}$$,i.e. $$\frac{3}{\sqrt{50}},\frac{4}{\sqrt{50}},\frac{-5}{\sqrt{50}}$$
Dividing the equation throughout by √50, we get
$$\frac{3}{\sqrt{50}} x+\frac{4}{\sqrt{50}} y-\frac{5}{\sqrt{50}} z=\frac{7}{\sqrt{50}}$$
The above equation is in the form of lx+my+nz=d, where d is the distance of the plane from the origin. So, the distance of the plane from the origin is $$\frac{7}{\sqrt{50}}$$.

8. Find the Cartesian equation of the plane $$\vec{r}.[(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}]$$=12.
a) (λ-μ)x+y+(3λ-2μ)z=12
b) (λ+3μ)x+(2+μ)y+(3λ-2μ)z=12
c) (λ+2μ)x-2λy+(3λ-2μ)z=12
d) (λ+2μ)x+(2λ-μ)y+(3λ-2μ)z=12

Explanation: Given that the equation of the plane is $$\vec{r}.[(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}]$$=12
We know that, $$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$$
∴$$(x\hat{i}+y\hat{j}+z\hat{k}).([(λ+2μ) \hat{i}+(2λ-μ) \hat{j}+(3λ-2μ)\hat{k}])$$=12
⇒(λ+2μ)x+(2λ-μ)y+(3λ-2μ)z=12 is the Cartesian equation of the plane.

9. Find the equation of the plane passing through the three points (2,2,0), (1,2,1), (-1,2,-2).
a) $$(\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]$$=0
b) $$(\vec{r}-(3\hat{i}-2\hat{k})).[(-\hat{i}+\hat{k})×(2\hat{i}-2\hat{j})]$$=0
c) $$(\vec{r}+(2\hat{i}+2\hat{j})).[(-\hat{i}-\hat{k})×(-3\hat{i}-2\hat{k})]$$=0
d) $$(\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}-\hat{k})×(3\hat{i}+2\hat{k})]$$=0

Explanation: Let $$\vec{a}=2\hat{i}+2\hat{j}, \,\vec{b}=\hat{i}+2\hat{j}+\hat{k}, \,\vec{c}=-\hat{i}+2\hat{j}-2\hat{k}$$
The vector equation of the plane passing through three points is given by
$$(\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]$$=0
$$(\vec{r}-(2\hat{i}+2\hat{j})).[((\hat{i}+2\hat{j}+\hat{k})-(2\hat{i}+2\hat{j}))×((-\hat{i}+2\hat{j}-2\hat{k})–(2\hat{i}+2\hat{j}))]$$=0
$$(\vec{r}-(2\hat{i}+2\hat{j})).[(-\hat{i}+\hat{k})×(-3\hat{i}-2\hat{k})]$$=0.

10. Find the Cartesian equation of the plane passing through the point (3,2,-3) and the normal to the plane is $$4\hat{i}-2\hat{j}+5\hat{k}$$?
a) 4x-2y+5z+7=0
b) 3x-2y-3z+1=0
c) 4x-y+5z+7=0
d) 4x-2y-z+7=0

Explanation: The position vector of the point (3,2,-3) is $$\vec{a}=3\hat{i}+2\hat{j}-3\hat{k}$$ and the normal vector $$\vec{N}$$ perpendicular to the plane is $$\vec{N}=4\hat{i}-2\hat{j}+5\hat{k}$$
Therefore, the vector equation of the plane is given by $$(\vec{r}-\vec{a}).\vec{N}$$=0
Hence, $$(\vec{r}-(3\hat{i}+2\hat{j}-3\hat{k})).(4\hat{i}-2\hat{j}+5\hat{k})$$=0
4(x-3)-2(y-2)+5(z+3)=0
4x-2y+5z+7=0.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

To practice all chapters and topics of class 12 Mathematics, here is complete set of 1000+ Multiple Choice Questions and Answers.