The angle between the planes 2x - y + z = 6 a | vedclass

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

Question

(D) The equation of the first plane is $2x - y + z = 6$,so its normal vector is $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$.The equation of the second

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) The equation of the first plane is $2x - y + z = 6$,so its normal vector is $\vec{n_1} = 2\hat{i} - \hat{j} + \hat{k}$.The equation of the second plane is $x + y + 2z = 7$,so its normal vector is $\vec{n_2} = \hat{i} + \hat{j} + 2\hat{k}$.The angle $	heta$ between two planes is given by $\cos 	heta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$.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 $|\vec{n_2}| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}$.Thus,$\cos 	heta = \frac{3}{\sqrt{6} \cdot \sqrt{6}} = \frac{3}{6} = \frac{1}{2}$.Therefore,$	heta = \cos^{-1}(\frac{1}{2}) = 60^o$.

The angle between the planes $2x - y + z = 6$ and $x + y + 2z = 7$ is ......... $^o$

Similar Questions

The plane $2x + 3y + 4z = 1$ meets the $X$-axis at $A$,the $Y$-axis at $B$,and the $Z$-axis at $C$. Then the centroid of $\triangle ABC$ is:

The equation $xy = 0$ in three-dimensional space represents

Let the plane passing through the point $(2,1,-1)$ and containing the line joining the points $(1,3,2)$ and $(1,2,1)$ make intercepts $p, q, r$ on the coordinate axes. Then $p+q+r=$

If $A$ and $B$ are the feet of the perpendiculars drawn from the point $Q(a, b, c)$ to the planes $YZ$ and $ZX$ respectively,then the equation of the plane passing through the points $A, B$,and $O$ is (where $O$ is the origin).

$A$ vector $\vec{n}$ is inclined to the $x$-axis at $45^\circ$,to the $y$-axis at $60^\circ$,and at an acute angle to the $z$-axis. If $\vec{n}$ is a normal to a plane passing through the point $(\sqrt{2}, -1, 1)$,then the equation of the plane is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module