Plane

1 Mark Questions

1. Write the vector equation of the plane passing through the point <math>(a, b, c)</math> and parallel to the

plane <math>\vec{r} \cdot (\hat{i} + \hat{i} + \hat{k}) = 2</math>. Delhi 2014

The required plane is passing through the point (a,b,c) whose position vector is <math>\vec{p} = a\hat{i} + b\hat{j} + c\hat{k}</math> and is parallel to the plane <math>\overrightarrow{r} \cdot (\widehat{i} + \widehat{i} + \widehat{k}) = 2.</math>

So, it is normal to the vector

<math>\vec{n} = \hat{i} + \hat{j} + \hat{k}</math>

Hence, required equation of plane is

<math>(\overrightarrow{r} - \overrightarrow{p}) \cdot \overrightarrow{n} = 0</math> <math>\Rightarrow</math> <math>\overrightarrow{r} \cdot \overrightarrow{n} = \overrightarrow{p} \cdot \overrightarrow{n}</math> <math display="block">\Rightarrow \overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = (a\hat{i} + b\hat{j} + c \hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k})</math> <math>\Rightarrow \overrightarrow{r} \cdot (\hat{i} + \hat{j} + \hat{k}) = a + b + c</math> (1)

1. Find the length of the perpendicular drawn from the origin to the plane <math>2x-3y+6z+21=0</math>. All India 2013

The distance from point <math>(x_1, y_1, z_1)</math> to the plane <math>Ax + By + Cz + D = 0</math> is <math display="block">\frac{Ax_1 + By_1 + Cz_1 + D}{\sqrt{A^2 + B^2 + C^2}}</math>.

Given, equation of plane is

<math>2x-3y+6z+21=0</math>

...(i)

:. Length of the perpendicular drawn from the origin to this plane

<math display="block">= \left| \frac{2 \cdot 0 - 3 \cdot 0 + 6 \cdot 0 + 21}{\sqrt{(2)^2 + (-3)^2 + (6)^2}} \right| = \left| \frac{0 - 0 + 0 + 21}{\sqrt{4 + 9 + 36}} \right|</math> <math>= 3 \text{ units}</math> (1)

1. Find distance of the plane <math>3x - 4y + 12z = 3</math>

from the origin. Delhi 2011

Given equation of plane is <math>3x - 4y + 12z - 3 = 0</math> and the point is <math>(0, 0, 0)</math>. We know that, distance of the plane <math>Ax + By + Cz + D = 0</math> from the point <math>(x_1, y_1, z_1)</math> is