Find the angle between the planes 2x - y + z | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The normal vectors to the planes $2x - y + z = 6$ and $x + y + 2z = 3$ are $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \

Vedclass · Accepted Answer

$\pi / 3$

Vedclass · Answer

(A) The normal vectors to the planes $2x - y + z = 6$ and $x + y + 2z = 3$ are $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$ respectively. The angle $	heta$ between the two planes is given by the formula $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$. Calculating the dot product: $\vec{n_1} \cdot \vec{n_2} = (2)(1) + (-1)(1) + (1)(2) = 2 - 1 + 2 = 3$. Calculating the magnitudes: $|\vec{n_1}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$ and $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$. Substituting these values into the formula: $\cos 	heta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$. Therefore,$	heta = \cos^{-1}(1/2) = \pi / 3$.

Find the angle between the planes $2x - y + z = 6$ and $x + y + 2z = 3$.

Similar Questions

Let $P$ be an arbitrary point such that the sum of the squares of the distances from the planes $x + y + z = 0$,$lx - nz = 0$,and $x - 2y + z = 0$ is equal to $9$. If the locus of the point $P$ is $x^2 + y^2 + z^2 = 9$,then the value of $l - n$ is equal to ......

The vector equation of a plane,which is at a distance of $8$ units from the origin and which is normal to the vector $\vec{n} = 2\hat{i} + \hat{j} + 2\hat{k}$,is

The angle between two planes is defined as:

Find the angle between the two planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$ using the vector method.

In each of the following cases,determine the direction cosines of the normal to the plane and the distance from the origin: $5y + 8 = 0$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module