Let a line L_1 pass through the origin and be | vedclass

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

Question

(A) The direction vectors of $L_2$ and $L_3$ are $\vec{d}_2 = (1, 2, 2)$ and $\vec{d}_3 = (2, 2, 1)$.
The direction vector of $L_1$ is $\vec{d}_1 = \

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(A) The direction vectors of $L_2$ and $L_3$ are $\vec{d}_2 = (1, 2, 2)$ and $\vec{d}_3 = (2, 2, 1)$.
The direction vector of $L_1$ is $\vec{d}_1 = \vec{d}_2 	imes \vec{d}_3 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 2 & 2 \ 2 & 2 & 1 \end{vmatrix} = \hat{i}(2-4) - \hat{j}(1-4) + \hat{k}(2-4) = (-2, 3, -2)$.
Since $L_1$ passes through the origin, its equation is $\vec{r} = k(-2, 3, -2)$.
For the intersection of $L_1$ and $L_2$: $(-2k, 3k, -2k) = (3+t, 2t-1, 2t+4)$.
$-2k = 3+t$, $3k = 2t-1$, $-2k = 2t+4$.
From $3+t = 2t+4$, we get $t = -1$. Then $-2k = 3-1 = 2$, so $k = -1$.
The intersection point $P$ is $(-2(-1), 3(-1), -2(-1)) = (2, -3, 2)$.
Let the point on $L_3$ be $Q = (3+2s, 3+2s, 2+s)$.
The distance $PQ = \sqrt{17}$, 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 $s = -10/9$.
Since $a \in Z$, we choose $s = -2$.
Then $Q = (3+2(-2), 3+2(-2), 2-2) = (-1, -1, 0)$.
Thus, $a = -1, b = -1, c = 0$.
$(a+b+c)^2 = (-1-1+0)^2 = (-2)^2 = 4$.

Similar Questions

The line $\frac{x + 3}{3} = \frac{y - 2}{-2} = \frac{z + 1}{1}$ and the plane $4x + 5y + 3z - 5 = 0$ intersect at a point

If the point of intersection of the lines $r = \hat{i} - 6\hat{j} + (p \sec \alpha) \hat{k} + t(\hat{i} + 2\hat{j} + \hat{k})$ and $r = 4\hat{j} + \hat{k} + \lambda(2\hat{i} + (p \tan \alpha) \hat{j} + 2\hat{k})$ is $8\hat{i} + 8\hat{j} + 9\hat{k}$, (where $0 < \alpha < \frac{\pi}{2}$), then $p =$

What is the reflection of the point $P(1, 3, 4)$ in the plane $2x - y + z + 3 = 0$?

The angle between the line $\frac{x-1}{2}=\frac{y+3}{1}=\frac{z+7}{2}$ and the plane $\bar{r} \cdot(6 \hat{\imath}-2 \hat{\jmath}-3 \hat{k})=5$ is

Find the angle between the line $\frac{x+1}{2}=\frac{y}{3}=\frac{z-3}{6}$ and the plane $10x+2y-11z=3$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module