Consider the lines L_1: x-1=y-2=z and L_2: x- | vedclass

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

Question

(B) The line $L_1$ is given by $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let $Q = (\lambda+1, \lambda+2, \lambda)$.Since $PQ \perp L_1$,the vec

Vedclass · Accepted Answer

$147$

Vedclass · Answer

(B) The line $L_1$ is given by $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-0}{1}$. Let $Q = (\lambda+1, \lambda+2, \lambda)$.Since $PQ \perp L_1$,the vector $\vec{PQ} = (\lambda+1-5, \lambda+2-1, \lambda-(-3)) = (\lambda-4, \lambda+1, \lambda+3)$ is perpendicular to the direction vector $\vec{v_1} = (1, 1, 1)$.Thus,$(\lambda-4)(1) + (\lambda+1)(1) + (\lambda+3)(1) = 0 \Rightarrow 3\lambda = 0 \Rightarrow \lambda = 0$.So,$Q = (1, 2, 0)$ and $\vec{PQ} = (-4, 1, 3)$.The line $L_2$ is given by $\frac{x-2}{1} = \frac{y-0}{1} = \frac{z-1}{1}$. Let $R = (\mu+2, \mu, \mu+1)$.Since $PR \perp L_2$,the vector $\vec{PR} = (\mu+2-5, \mu-1, \mu+1-(-3)) = (\mu-3, \mu-1, \mu+4)$ is perpendicular to the direction vector $\vec{v_2} = (1, 1, 1)$.Thus,$(\mu-3)(1) + (\mu-1)(1) + (\mu+4)(1) = 0 \Rightarrow 3\mu = 0 \Rightarrow \mu = 0$.So,$R = (2, 0, 1)$ and $\vec{PR} = (-3, -1, 4)$.The area of $	riangle PQR$ is $A = \frac{1}{2} |\vec{PQ} 	imes \vec{PR}|$.$\vec{PQ} 	imes \vec{PR} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ -4 & 1 & 3 \ -3 & -1 & 4 \end{vmatrix} = \hat{i}(4 - (-3)) - \hat{j}(-16 - (-9)) + \hat{k}(4 - (-3)) = 7\hat{i} + 7\hat{j} - 7\hat{k}$.$|\vec{PQ} 	imes \vec{PR}| = \sqrt{7^2 + 7^2 + (-7)^2} = \sqrt{49 	imes 3} = 7\sqrt{3}$.$A = \frac{1}{2} 	imes 7\sqrt{3} = \frac{7\sqrt{3}}{2}$.$4A^2 = 4 	imes \frac{49 	imes 3}{4} = 147$.

Consider the lines $L_1: x-1=y-2=z$ and $L_2: x-2=y=z-1$. Let the feet of the perpendiculars from the point $P(5,1,-3)$ on the lines $L_1$ and $L_2$ be $Q$ and $R$ respectively. If the area of the triangle $PQR$ is $A$,then $4A^2$ is equal to:

Similar Questions

If the lines $\frac{x-3}{2}=\frac{y-2}{3}=\frac{z-1}{\lambda}$ and $\frac{x-2}{3}=\frac{y-3}{2}=\frac{z-2}{3}$ are coplanar,then $\sin ^{-1}(\sin \lambda)+\cos ^{-1}(\cos \lambda)=$

If the lines given by $\bar{r} = 2 \hat{i} + \lambda(\hat{i} + 2 \hat{j} + m \hat{k})$ and $\bar{r} = \hat{i} + \mu(2 \hat{i} + \hat{j} + 6 \hat{k})$ are perpendicular,then the value of $m$ is:

The point of intersection of the lines $\frac{x - 5}{3} = \frac{y - 7}{-1} = \frac{z + 2}{1}$ and $\frac{x + 3}{-36} = \frac{y - 3}{2} = \frac{z - 6}{4}$ is

The Cartesian equation of the line which is parallel to the vector $3 \hat{i} + 2 \hat{j} - 8 \hat{k}$ and passes through the point $(5, 2, -4)$ is . . . . . . .

Let $P$ and $Q$ be the points on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ which are at a distance of $6$ units from the point $R(1,2,3)$. If the centroid of the triangle $PQR$ is $(\alpha, \beta, \gamma)$,then $\alpha^2+\beta^2+\gamma^2$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module