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 the plane which passes through $(2, -3, 1)$ and is normal to the line joining the points $(3, 4, -1)$ and $(2, -1, 5)$ is given by:

Find the angle between the planes $\vec{r} \cdot (2\hat{i} - \hat{j} + \hat{k}) = 6$ and $\vec{r} \cdot (\hat{i} + \hat{j} + 2\hat{k}) = 5$.

The equation of a plane containing the point $(1, -1, 2)$ and perpendicular to the planes $2x + 3y - 2z = 5$ and $x + 2y - 3z = 8$ is

The equation of the plane passing through $(4,4,0)$ and perpendicular to the planes $2x+y+2z+3=0$ and $3x+3y+2z-8=0$ is

The vector equation of a plane which is at a distance of $5 \text{ units}$ from the origin and normal to the vector $\vec{n} = 2\hat{i} + \hat{j} - 2\hat{k}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module