A bisector of the angle between the normals o | vedclass

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

Question

(D) The normals to the given planes $4x + 3y - 5 = 0$ and $x + 2y + 2z - 4 = 0$ are $\vec{n_1} = 4\hat{i} + 3\hat{j} + 0\hat{k}$ and $\vec{n_2} = \hat

Vedclass · Accepted Answer

$7\hat{i} - \hat{j} - 10\hat{k}$

Vedclass · Answer

(D) The normals to the given planes $4x + 3y - 5 = 0$ and $x + 2y + 2z - 4 = 0$ are $\vec{n_1} = 4\hat{i} + 3\hat{j} + 0\hat{k}$ and $\vec{n_2} = \hat{i} + 2\hat{j} + 2\hat{k}$.To find the bisector of the angle between these normals,we first normalize the vectors:$|\vec{n_1}| = \sqrt{4^2 + 3^2 + 0^2} = 5$$|\vec{n_2}| = \sqrt{1^2 + 2^2 + 2^2} = 3$The unit vectors are $\hat{n_1} = \frac{4}{5}\hat{i} + \frac{3}{5}\hat{j}$ and $\hat{n_2} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} + \frac{2}{3}\hat{k}$.The angle bisector is along the vector $\hat{n_1} \pm \hat{n_2}$.Taking the sum: $(\frac{4}{5} + \frac{1}{3})\hat{i} + (\frac{3}{5} + \frac{2}{3})\hat{j} + \frac{2}{3}\hat{k} = \frac{17}{15}\hat{i} + \frac{19}{15}\hat{j} + \frac{10}{15}\hat{k}$,which is proportional to $17\hat{i} + 19\hat{j} + 10\hat{k}$.Taking the difference: $(\frac{4}{5} - \frac{1}{3})\hat{i} + (\frac{3}{5} - \frac{2}{3})\hat{j} - \frac{2}{3}\hat{k} = \frac{7}{15}\hat{i} - \frac{1}{15}\hat{j} - \frac{10}{15}\hat{k}$,which is proportional to $7\hat{i} - \hat{j} - 10\hat{k}$.Comparing with the options,$7\hat{i} - \hat{j} - 10\hat{k}$ is the correct vector.

$A$ bisector of the angle between the normals of the planes $4x + 3y = 5$ and $x + 2y + 2z = 4$ is along the vector

Similar Questions

The equation of a plane containing the point $(1, -1, 1)$ and parallel to the plane $2x + 3y - 4z = 17$ is

If for some $\alpha$ and $\beta$ in $\mathbb{R},$ the intersection of the following three planes $x+4y-2z=1$,$x+7y-5z=\beta$,and $x+5y+\alpha z=5$ is a line in $\mathbb{R}^{3},$ then $\alpha+\beta$ is equal to

If the equation of the plane passing through the point $(3,2,5)$ and perpendicular to the planes $2x-3y+5z=7$ and $5x+2y-3z=11$ is $x+by+cz+d=0$,then $2b+3c+d=$

The coordinates of the foot of the perpendicular drawn from the origin to the plane $2x + y - 2z = 18$ are

In space,the equation $by + cz + d = 0$ represents a plane perpendicular to the plane:

Vedclass Test Series

Exam Paper Generator

Online Exam Module