Let the lines L_1: vec{r}=hat{i}+2hat{j}+3hat | vedclass

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

Question

(B) To find the point of intersection $R$,we equate the expressions for $\vec{r}$:$2\lambda+1 = 5\mu+4$,$3\lambda+2 = 2\mu+1$,$4\lambda+3 = \mu$.Solvi

Vedclass · Accepted Answer

$360$

Vedclass · Answer

(B) To find the point of intersection $R$,we equate the expressions for $\vec{r}$:$2\lambda+1 = 5\mu+4$,$3\lambda+2 = 2\mu+1$,$4\lambda+3 = \mu$.Solving these,we get $\lambda = -1$ and $\mu = -1$.Thus,the point $R$ is $(-1, -1, -1)$.Point $P$ lies on $L_1$,so $P = (2\lambda+1, 3\lambda+2, 4\lambda+3)$.Given $|\overrightarrow{PR}| = \sqrt{29}$,so $PR^2 = 29$.$(2\lambda+1 - (-1))^2 + (3\lambda+2 - (-1))^2 + (4\lambda+3 - (-1))^2 = 29$.$(2\lambda+2)^2 + (3\lambda+3)^2 + (4\lambda+4)^2 = 29$.$29(\lambda+1)^2 = 29 \Rightarrow (\lambda+1)^2 = 1 \Rightarrow \lambda = 0$ or $\lambda = -2$.If $\lambda = 0$,$P = (1, 2, 3)$ (First octant). If $\lambda = -2$,$P = (-3, -4, -5)$ (Not in first octant).So,$P = (1, 2, 3)$.Point $Q$ lies on $L_2$,so $Q = (5\mu+4, 2\mu+1, \mu)$.Given $|\overrightarrow{PQ}|^2 = \frac{47}{3}$.$(5\mu+4-1)^2 + (2\mu+1-2)^2 + (\mu-3)^2 = \frac{47}{3}$.$(5\mu+3)^2 + (2\mu-1)^2 + (\mu-3)^2 = \frac{47}{3}$.$25\mu^2 + 30\mu + 9 + 4\mu^2 - 4\mu + 1 + \mu^2 - 6\mu + 9 = \frac{47}{3}$.$30\mu^2 + 20\mu + 19 = \frac{47}{3} \Rightarrow 90\mu^2 + 60\mu + 57 = 47 \Rightarrow 90\mu^2 + 60\mu + 10 = 0$.$9\mu^2 + 6\mu + 1 = 0 \Rightarrow (3\mu+1)^2 = 0 \Rightarrow \mu = -\frac{1}{3}$.$Q = (5(-\frac{1}{3})+4, 2(-\frac{1}{3})+1, -\frac{1}{3}) = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})$.$(QR)^2 = (\frac{7}{3} - (-1))^2 + (\frac{1}{3} - (-1))^2 + (-\frac{1}{3} - (-1))^2 = (\frac{10}{3})^2 + (\frac{4}{3})^2 + (\frac{2}{3})^2 = \frac{100+16+4}{9} = \frac{120}{9}$.$27(QR)^2 = 27 	imes \frac{120}{9} = 3 	imes 120 = 360$.

Similar Questions

For a positive real number $p$,if the perpendicular distance from a point $-\hat{i} + p\hat{j} - 3\hat{k}$ to the plane $\vec{r} \cdot (2\hat{i} - 3\hat{j} + 6\hat{k}) = 7$ is $6$ units,then $p=$

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

Let $Q$ be the foot of the perpendicular drawn from the point $P(1, 2, 3)$ to the plane $x + 2y + z = 14$. If $R$ is a point on the plane such that $\angle PRQ = 60^{\circ}$,then the area of $\triangle PQR$ is equal to:

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

The coordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module