The Plane (VSAQs)
Maths-1B | 7. The Plane – VSAQs:
Welcome to VSAQs in Chapter 7: The Plane. This page contains the most important VSAQs in this chapter. Aim to secure top marks in your exams by understanding these clear and straightforward Very Short Answer Questions.
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
$$(x_1, y_1, z_1) = (1, 2, -3), a = 2, b = -3, c = 6$$
$$\text{The equation of the plane passing through }(x_1, y_1, z_1)\text{ and parallel to the plane }ax + by + cz + d = 0\text{ is}$$
$$a(x – x_1) + b(y – y_1) + c(z – z_1) = 0$$
$$\Rightarrow 2(x – 1) – 3(y – 2) + 6(z + 3) = 0$$
$$\Rightarrow 2x – 2 – 3y + 6 + 6z + 18 = 0$$
$$\Rightarrow 2x – 3y + 6z + 22 = 0$$