Let S be the reflection of a point Q with res | vedclass

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

Question

(D) The equation of the plane is $\vec{r} = \hat{k} + t(-\hat{i} + \hat{j}) + p(-\hat{i} + \hat{k})$.This represents a plane passing through $(0, 0, 1

Vedclass · Accepted Answer

$A, B, C$

Vedclass · Answer

(D) The equation of the plane is $\vec{r} = \hat{k} + t(-\hat{i} + \hat{j}) + p(-\hat{i} + \hat{k})$.This represents a plane passing through $(0, 0, 1)$ with normal vector $\vec{n} = (-\hat{i} + \hat{j}) 	imes (-\hat{i} + \hat{k}) = \hat{i} + \hat{j} + \hat{k}$.The equation of the plane is $1(x-0) + 1(y-0) + 1(z-1) = 0$,which simplifies to $x + y + z = 1$.Given $Q = (10, 15, 20)$ and $S = (\alpha, \beta, \gamma)$,the reflection formula is $\frac{\alpha-10}{1} = \frac{\beta-15}{1} = \frac{\gamma-20}{1} = -2 \frac{10+15+20-1}{1^2+1^2+1^2} = -2 \frac{44}{3} = -\frac{88}{3}$.Thus,$\alpha = 10 - \frac{88}{3} = -\frac{58}{3}$,$\beta = 15 - \frac{88}{3} = -\frac{43}{3}$,and $\gamma = 20 - \frac{88}{3} = -\frac{28}{3}$.Checking the options:$(A)$ $3(\alpha+\beta) = 3(-\frac{58}{3} - \frac{43}{3}) = -101$ (True).$(B)$ $3(\beta+\gamma) = 3(-\frac{43}{3} - \frac{28}{3}) = -71$ (True).$(C)$ $3(\gamma+\alpha) = 3(-\frac{28}{3} - \frac{58}{3}) = -86$ (True).$(D)$ $3(\alpha+\beta+\gamma) = 3(-\frac{58+43+28}{3}) = -129$ (False).Therefore,options $A, B, C$ are correct.

Similar Questions

The equation of the plane passing through the intersection of the planes $x + y + z = 6$ and $2x + 3y + 4z + 5 = 0$ and the point $(1, 1, 1)$ is:

If the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 2}{4}$ meets the plane $x + 2y + 3z = 15$ at a point $P$,then the distance of $P$ from the origin is

The equation of the plane through the intersection of the planes $x+y+z=1$ and $2x+3y-z+4=0$ and parallel to the $x$-axis is:

The foot of the perpendicular from a point on the circle $x^{2} + y^{2} = 1, z = 0$ to the plane $2x + 3y + z = 6$ lies on which one of the following curves?

The equation of the plane containing the straight line $\frac{x}{3}=\frac{y}{2}=\frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{4}=\frac{y}{3}=\frac{z}{2}$ and $\frac{x}{2}=\frac{y}{-4}=\frac{z}{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module