9 Most VSAQ’s of The Plane Chapter in Inter 1st Year Maths-1B (TS/AP)

Table of Contents

2 Marks

VSAQ-1 : Write the equation of the plane 4x – 4y + 2z + 5 = 0 in the intercept form

The given equation of the plane is

$$4x – 4y + 2z + 5 = 0$$

$$\Rightarrow 4x – 4y + 2z = -5$$

$$\Rightarrow \frac{4x}{-5} + \frac{-4y}{-5} + \frac{2z}{-5} = 1$$

$$\Rightarrow \frac{x}{-5/4} + \frac{y}{5/4} + \frac{z}{-5/2} = 1$$

which is in the intercept form: $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

VSAQ-2 : Find the intercepts of the plane 4x + 3y – 2z + 2 = 0 on the coordinate axes

The given equation of the plane is

$$4x + 3y – 2z + 2 = 0$$

$$\Rightarrow 4x + 3y – 2z = -2$$

$$\Rightarrow \frac{4x}{-2} + \frac{3y}{-2} – \frac{2z}{-2} = 1$$

$$\Rightarrow \frac{x}{-1/2} + \frac{y}{-2/3} + \frac{z}{1} = 1$$

Which is in the intercept form $$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

Hence, the intercepts are:

$$x\text{-intercept} = -1/2$$

$$y\text{-intercept} = -2/3$$

$$z\text{-intercept} = 1$$

VSAQ-3 : Find the equation of the plane which makes intercepts 1,2,4 on the x,y,z-axes respectively

The equation of the plane with intercepts a,b,c is

$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

The required equation of the plane is

$$\frac{x}{1} + \frac{y}{2} + \frac{z}{4} = 1$$

$$\Rightarrow 4x + 2y + \frac{z}{4} = 1$$

$$\Rightarrow 4x + 2y + z = 4$$

VSAQ-4 : Reduce the equation x + 2y – 3z – 6 = 0 of the plane to the normal form

Equation of the plane is

$$x + 2y – 3z – 6 = 0$$

$$\Rightarrow x + 2y – 3z = 6$$

Dividing the above equation by

$$\sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$$

We have:

$$\frac{1}{\sqrt{14}}x + \frac{2}{\sqrt{14}}y + \frac{-3}{\sqrt{14}}z = \frac{6}{\sqrt{14}}$$

Which is in the normal form

VSAQ-5 : Find a traid of d.c’s of the normal to the plane x + 2y + 2z – 4 = 0

The d.c’s of the plane ax + by + cz + d = 0 are

$$\pm\left(\frac{a}{\sqrt{a^2 + b^2 + c^2}}, \frac{b}{\sqrt{a^2 + b^2 + c^2}}, \frac{c}{\sqrt{a^2 + b^2 + c^2}}\right)$$

The direction cosines (d.c.’s) of the plane 𝑥 + 2𝑦 + 2𝑧 − 4 = 0 are:

$$\pm\left(\frac{1}{\sqrt{1 + 4 + 4}}, \frac{2}{\sqrt{1 + 4 + 4}}, \frac{2}{\sqrt{1 + 4 + 4}}\right) = \pm\left(\frac{1}{3}, \frac{2}{3}, \frac{2}{3}\right)$$

VSAQ-6 : Find the angle between the planes 2x – y + z = 6 and x + y + 2z = 7

Given planes are

$$2x – y + z = 6, x + y + 2z = 7$$

$$\Rightarrow a_1 = 2, b_1 = -1, c_1 = 1; a_2 = 1, b_2 = 1, c_2 = 2$$

$$\cos \theta = \left|\frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}\right|$$

$$= \left|\frac{2 \cdot 1 – 1 \cdot 1 + 1 \cdot 2}{\sqrt{2^2 + 1^2 + 1^2} \cdot \sqrt{1^2 + 1^2 + 2^2}}\right|$$

$$= \left|\frac{2 – 1 + 2}{\sqrt{4 + 1 + 1} \cdot \sqrt{1 + 1 + 4}}\right| = \left|\frac{3}{\sqrt{6} \cdot \sqrt{6}}\right| = \frac{3}{6} = \frac{1}{2}$$

$$\cos \theta = \frac{1}{2} = \cos 60^\circ$$

$$\Rightarrow \theta = 60^\circ$$

VSAQ-7 : Find the angle between the planes x + 2y + 2z – 5 = 0 and 3x + 3y 2z – 8 = 0

Given planes are

$$x + 2y + 2z – 5 = 0, 3x + 3y + 2z – 8 = 0$$

$$\Rightarrow a_1 = 1, b_1 = 2, c_1 = 2; a_2 = 3, b_2 = 3, c_2 = 2$$

$$\cos \theta = \left|\frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2 + b_1^2 + c_1^2} \cdot \sqrt{a_2^2 + b_2^2 + c_2^2}}\right|$$

$$= \left|1 \cdot 3 + 2 \cdot 3 + 2 \cdot 2\right| / \sqrt{1 + 4 + 4} \sqrt{9 + 9 + 4}$$

$$= \frac{13}{\sqrt{9} \cdot \sqrt{22}} = \frac{13}{3\sqrt{22}}$$

$$\cos \theta = \frac{13}{3\sqrt{22}}$$

$$\Rightarrow \theta = \cos^{-1}\left(\frac{13}{3\sqrt{22}}\right)$$

VSAQ-8 : Find the equation of the plane passing through the point (1, 1, 1) and parallel to the plane x + 2y + 3z – 7 = 0

Here $$(x_1, y_1, z_1) = (1, 1, 1), a = 1, b = 2, c = 3$$

$$\text{The equation of the plane passing through }(x_1, y_1, z_1)\text{ and parallel to the plane }ax + by + cz + d = 0\text{ is}$$

$$a(x – x_1) + b(y – y_1) + c(z – z_1) = 0$$

$$\Rightarrow 1(x – 1) + 2(y – 1) + 3(z – 1) = 0$$

$$\Rightarrow x – 1 + 2y – 2 + 3z – 3 = 0$$

$$\Rightarrow x + 2y + 3z – 6 = 0$$

VSAQ-9 : Find the equation of the plane passing through the point (1, 2, -3) and parallel to the plane 2x – 3y + 6z = 0

Here

$$(x_1, y_1, z_1) = (1, 2, -3), a = 2, b = -3, c = 6$$

$$\text{The equation of the plane passing through }(x_1, y_1, z_1)\text{ and parallel to the plane }ax + by + cz + d = 0\text{ is}$$

$$a(x – x_1) + b(y – y_1) + c(z – z_1) = 0$$

$$\Rightarrow 2(x – 1) – 3(y – 2) + 6(z + 3) = 0$$

$$\Rightarrow 2x – 2 – 3y + 6 + 6z + 18 = 0$$

$$\Rightarrow 2x – 3y + 6z + 22 = 0$$