Find the equation of the planes bisecting the | vedclass

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

Question

(D) The equations of the planes are $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3

Vedclass · Accepted Answer

$\vec{r} \cdot (25\hat{i} + 17\hat{j} + 62\hat{k}) = 238$

Vedclass · Answer

(D) The equations of the planes are $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19 = 0$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$.The equation of the angle bisector planes is given by:$\frac{\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19}{\sqrt{1^2 + 2^2 + 2^2}} = \pm \frac{\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3}{\sqrt{4^2 + (-3)^2 + 12^2}}$$\frac{\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19}{3} = \pm \frac{\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3}{13}$Case $1$ (Positive sign):$13(\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19) = 3(\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3)$$\vec{r} \cdot (13\hat{i} + 26\hat{j} + 26\hat{k} - 12\hat{i} + 9\hat{j} - 36\hat{k}) = 9 + 247$$\vec{r} \cdot (\hat{i} + 35\hat{j} - 10\hat{k}) = 256$Case $2$ (Negative sign):$13(\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) - 19) = -3(\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3)$$\vec{r} \cdot (13\hat{i} + 26\hat{j} + 26\hat{k} + 12\hat{i} - 9\hat{j} + 36\hat{k}) = -9 + 247$$\vec{r} \cdot (25\hat{i} + 17\hat{j} + 62\hat{k}) = 238$Comparing with the options,the correct equation is $\vec{r} \cdot (25\hat{i} + 17\hat{j} + 62\hat{k}) = 238$.

Find the equation of the planes bisecting the angles between the planes $\vec{r} \cdot (\hat{i} + 2\hat{j} + 2\hat{k}) = 19$ and $\vec{r} \cdot (4\hat{i} - 3\hat{j} + 12\hat{k}) + 3 = 0$.

Similar Questions

The vertices of a tetrahedron are $O(0, 0, 0)$,$A(1, 2, 1)$,$B(2, 1, 3)$,and $C(-1, 1, 2)$. Find the angle between the faces $OAB$ and $ABC$.

The equation of the plane passing through the three points $(1, 1, 1)$,$(1, -1, 1)$,and $(-7, -3, -5)$ is:

If the length of the perpendicular drawn from the origin to a plane is $7$ units and its direction ratios are $-3, 2, 6$,then the equation of the plane is:

The values of $a$ for which the two points $(1, a, 1)$ and $(-3, 0, a)$ lie on the opposite sides of the plane $3x + 4y - 12z + 13 = 0$ satisfy:

$A$ plane meets the coordinate axes at $A, B, C$ such that the centroid of the triangle $ABC$ is $(1, 2, 4)$. Then,the equation of the plane is

Vedclass Test Series

Exam Paper Generator

Online Exam Module