Find the equation of the line passing through | vedclass

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

Question

(D) Let the given line be $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1} = k$. Any point $Q$ on $L$ is $(2k - 1, 3k + 3, -k - 2)$.The direct

Vedclass · Accepted Answer

$\frac{x - 2}{-24} = \frac{y + 3}{13} = \frac{z - 1}{-9}$

Vedclass · Answer

(D) Let the given line be $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1} = k$. Any point $Q$ on $L$ is $(2k - 1, 3k + 3, -k - 2)$.The direction ratios of $PQ$ are $(2k - 1 - 2, 3k + 3 - (-3), -k - 2 - 1) = (2k - 3, 3k + 6, -k - 3)$.Since $PQ \perp L$, the dot product of the direction ratios of $PQ$ and $L$ is zero:$2(2k - 3) + 3(3k + 6) - 1(-k - 3) = 0$.$4k - 6 + 9k + 18 + k + 3 = 0 \implies 14k + 15 = 0 \implies k = -\frac{15}{14}$.The direction ratios of $PQ$ are $(2(-\frac{15}{14}) - 3, 3(-\frac{15}{14}) + 6, -(-\frac{15}{14}) - 3) = (-\frac{72}{14}, \frac{39}{14}, -\frac{27}{14})$.Multiplying by $-\frac{14}{3}$, we get the direction ratios as $(24, -13, 9)$.Wait, checking the dot product again: $2(24) + 3(-13) - 1(9) = 48 - 39 - 9 = 0$. The vector is $(24, -13, 9)$.The line equation is $\frac{x - 2}{24} = \frac{y + 3}{-13} = \frac{z - 1}{9}$.Comparing with options, option $A$ is $\frac{x - 2}{24} = \frac{y + 3}{13} = \frac{z - 1}{9}$. Re-evaluating the direction ratios: $(2k-3, 3k+6, -k-3)$. For $k=-15/14$, $DRs = (-72/14, 39/14, -27/14) \propto (-24, 13, -9)$.Thus, the equation is $\frac{x - 2}{-24} = \frac{y + 3}{13} = \frac{z - 1}{-9}$, which matches option $D$.

Find the equation of the line passing through the point $P(2, -3, 1)$ and perpendicular to the line $L: \frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z + 2}{-1}$.

Similar Questions

The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

Lines $\frac{x-5}{7}=\frac{y-5}{k}=\frac{z-2}{1}$ and $\frac{x}{1}=\frac{y-3}{2}=\frac{z+1}{3}$ are perpendicular to each other,then the value of $k=$ . . . . . . .

Let $L$ be the line $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+3}{6}$ and let $S$ be the set of all points $(a, b, c)$ on $L$, whose distance from the point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

Find the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$.

If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module