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 vector equation of the plane passing through the point $(2, 1, -1)$ and the line of intersection of the planes $r \cdot (i + 3j - k) = 0$ and $r \cdot (j + 2k) = 0$ is:

If the line $\frac{x - x_1}{l} = \frac{y - y_1}{m} = \frac{z - z_1}{n}$ is parallel to the plane $ax + by + cz + d = 0$,then which of the following is true?

For the line $\frac{x + 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{3}$, identify the incorrect statement among the following.

The ratio in which the line segment joining the points $(2, -4, 3)$ and $(-4, 5, -6)$ is divided by the plane $3x + 2y + z - 4 = 0$ is:

Let the points $P, Q$ and $R$ have position vectors $\overrightarrow{r_1} = 3i - 2j - k, \overrightarrow{r_2} = i + 3j + 4k$ and $\overrightarrow{r_3} = 2i + j - 2k$ respectively relative to an origin $O$. Then the distance of $P$ from the plane $OQR$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module