Let P (-2,-1,1) and Q left(frac{56}{17}, frac | vedclass

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

Question

(C) In a rhombus $PRQS$,the diagonals $PQ$ and $RS$ are perpendicular to each other.The direction ratios of diagonal $PQ$ are given by $(x_Q - x_P, y_

Vedclass · Accepted Answer

$450$

Vedclass · Answer

(C) In a rhombus $PRQS$,the diagonals $PQ$ and $RS$ are perpendicular to each other.The direction ratios of diagonal $PQ$ are given by $(x_Q - x_P, y_Q - y_P, z_Q - z_P) = \left(\frac{56}{17} + 2, \frac{43}{17} + 1, \frac{111}{17} - 1\right) = \left(\frac{90}{17}, \frac{60}{17}, \frac{94}{17}\right)$.Since $PQ \perp RS$,the dot product of their direction ratios must be zero.Let the direction ratios of $RS$ be $(\alpha, -1, \beta)$. Then,$\frac{90}{17}(\alpha) + \frac{60}{17}(-1) + \frac{94}{17}(\beta) = 0$.Multiplying by $17$,we get $90\alpha - 60 + 94\beta = 0$,which simplifies to $90\alpha + 94\beta = 60$,or $45\alpha + 47\beta = 30$.We need to find integers $\alpha$ and $\beta$ with minimum absolute values satisfying $47\beta = 30 - 45\alpha$.Testing values: If $\alpha = -15$,then $47\beta = 30 - 45(-15) = 30 + 675 = 705$. Thus,$\beta = \frac{705}{47} = 15$.Wait,checking the equation $45\alpha + 47\beta = 30$: If $\alpha = -15$,$45(-15) + 47\beta = 30 \Rightarrow -675 + 47\beta = 30 \Rightarrow 47\beta = 705 \Rightarrow \beta = 15$.Thus,$\alpha^2 + \beta^2 = (-15)^2 + 15^2 = 225 + 225 = 450$.

Similar Questions

If the line $\frac{x - 3}{1} = \frac{y + 2}{-1} = \frac{z + \lambda}{-2}$ lies in the plane $2x - 4y + 3z = 2$,then the shortest distance between this line and the line $\frac{x - 1}{12} = \frac{y}{9} = \frac{z}{4}$ is

The equation of the plane passing through the origin and perpendicular to the line $x = 2y = 3z$ is

Let $P$ be the plane passing through the line $\frac{x-1}{1}=\frac{y-2}{-3}=\frac{z+5}{7}$ and the point $(2,4,-3)$. If the image of the point $(-1,3,4)$ in the plane $P$ is $(\alpha, \beta, \gamma)$,then $\alpha+\beta+\gamma$ is equal to

The image of the point with position vector $(\hat{i}+3 \hat{j}+4 \hat{k})$ in the plane $r \cdot(2 \hat{i}-\hat{j}+\hat{k})+3=0$ is

The distance of the point $2i + j - k$ from the plane $r \cdot (i - 2j + 4k) = 9$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module