A line l passing through the origin is perpen | vedclass

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

Question

(C) The direction ratios of $l_{1}$ are $\vec{v}_{1} = (1, 2, 2)$ and for $l_{2}$ are $\vec{v}_{2} = (2, 2, 1)$.Since line $l$ is perpendicular to bot

Vedclass · Accepted Answer

$44$

Vedclass · Answer

(C) The direction ratios of $l_{1}$ are $\vec{v}_{1} = (1, 2, 2)$ and for $l_{2}$ are $\vec{v}_{2} = (2, 2, 1)$.Since line $l$ is perpendicular to both $l_{1}$ and $l_{2}$,its direction vector is $\vec{v} = \vec{v}_{1} 	imes \vec{v}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = -2\hat{i} + 3\hat{j} - 2\hat{k}$.Since $l$ passes through the origin,its equation is $\vec{r} = \mu(-2\hat{i} + 3\hat{j} - 2\hat{k})$.For the intersection of $l$ and $l_{1}$:$3+t = -2\mu$,$-1+2t = 3\mu$,$4+2t = -2\mu$.Solving these,we get $t = -1$ and $\mu = -1$. The point of intersection $P$ is $(2, -3, 2)$.Let $Q$ be a point on $l_{2}$,$Q = (3+2s, 3+2s, 2+s)$.Given $PQ = \sqrt{17}$,so $PQ^{2} = 17$.$(3+2s-2)^{2} + (3+2s+3)^{2} + (2+s-2)^{2} = 17$.$(2s+1)^{2} + (2s+6)^{2} + s^{2} = 17$.$4s^{2} + 4s + 1 + 4s^{2} + 24s + 36 + s^{2} = 17$.$9s^{2} + 28s + 20 = 0$.$(9s+10)(s+2) = 0$,so $s = -2$ or $s = -10/9$.For $s = -10/9$,$Q = (3-20/9, 3-20/9, 2-10/9) = (7/9, 7/9, 8/9)$,which is in the first octant.Thus,$a=7/9, b=7/9, c=8/9$.$18(a+b+c) = 18(7/9 + 7/9 + 8/9) = 18(22/9) = 44$.

Similar Questions

For non-coplanar vectors $a, b$ and $c$, if the point of intersection of the line $r=a+t(b-c)$ and the plane $r=b+c+x(a-b)+y(c+a)$ is $l a+m b+n c$, then $3 l+4 m+2 n=$

If a plane $x+y+z-5=0$ intersects the line joining $A(1,1,1)$ and $B(2,2,2)$ at $P$, then $AP: PB=$

The position vectors of points $A$ and $B$ are $i - j + 3k$ and $3i + 3j + 3k$ respectively. The equation of a plane is $r \cdot (5i + 2j - 7k) + 9 = 0$. The points $A$ and $B$:

The line $\frac{x - 2}{3} = \frac{y - 3}{4} = \frac{z - 4}{5}$ is parallel to which of the following planes?

If $\frac{x-4}{1}=\frac{y-2}{1}=\frac{z-7}{2}$ lies in the plane $ax+by+z=7$, then $a+b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module