Name: Equation of Plane Making Intercepts on Axes
Uploaded: 2019-07-19
Duration: 3 min 49 s
Description: In 2-D geometry, you have studied the Intercept form of the Equation of the straight line. Could you think of the Intercept form of the equation of a plane?...

Question

Find the angle between the planes whose vector equations are $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 and \vec{r} \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$ .

Accepted Answer

Given that, the equations of the planes in vector form are: $\vec{r} \cdot (2 \hat{i} + 2 \hat{j} - 3 \hat{k}) = 5 and \vec{r} \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k}) = 3$

The angle between two planes is equal to the angle between their normals. So, the angle between two planes given in the vector form, $\vec{r} \cdot \vec{n_{1}} = d_{1}$ and $\vec{r} \cdot \vec{n_{2}} = d_{2}$ is $\cos θ = |\frac{\vec{n_{1}} \cdot \vec{n_{2}}}{|\vec{n_{1}}| |\vec{n_{2}}|}|$ .

From the question, we get $\vec{n_{1}} = 2 \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\vec{n_{2}} = 3 \hat{i} - 3 \hat{j} + 5 \hat{k}$ . Therefore, the angle can be computed as $\cos θ = |\frac{(2 \hat{i} + 2 \hat{j} - 3 \hat{k}) \cdot (3 \hat{i} - 3 \hat{j} + 5 \hat{k})}{|2 \hat{i} + 2 \hat{j} - 3 \hat{k}| |3 \hat{i} - 3 \hat{j} + 5 \hat{k}|}|$

$\Rightarrow \cos θ = |\frac{(2 \times 3) + (2 \times - 3) + (- 3 \times 5)}{\sqrt{2^{2} + 2^{2} + (- 3)^{2}} \times \sqrt{3^{2} + (- 3)^{2} + 5^{2}}}|$

$\Rightarrow \cos θ = |\frac{6 - 6 - 15}{\sqrt{17} \times \sqrt{43}}|$

$\Rightarrow \cos θ = |\frac{- 15}{\sqrt{731}}|$

$\Rightarrow θ = \cos^{- 1} (\frac{15}{\sqrt{731}})$

Hence, the angle between given two planes is $\cos^{- 1} (\frac{15}{\sqrt{731}})$ .

Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Karnataka Board-2019

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Karnataka Board-2019

Questions from Embibe Experts Solutions for Chapter: Three Dimensional Geometry, Exercise 1: Karnataka Board-2019 with Hints & Solutions

Learn and Prepare for any exam you want !