If the straight lines vec{r} = (1, 2, 3) + k( | vedclass

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

Question

(A) Let the two lines be $L_1: (1 + k\lambda, 2 + 2k, 3 + 3k)$ and $L_2: (2 + 3m, 3 + m\lambda, 1 + 2m)$.Since the lines intersect,there exist $k$ and

Vedclass · Accepted Answer

$-5$

Vedclass · Answer

(A) Let the two lines be $L_1: (1 + k\lambda, 2 + 2k, 3 + 3k)$ and $L_2: (2 + 3m, 3 + m\lambda, 1 + 2m)$.Since the lines intersect,there exist $k$ and $m$ such that:$1 + k\lambda = 2 + 3m$  $(1)$$2 + 2k = 3 + m\lambda$  $(2)$$3 + 3k = 1 + 2m$  $(3)$From $(3)$,$3k - 2m = -2 \implies m = \frac{3k + 2}{2}$.Substitute $m$ into $(2)$: $2 + 2k = 3 + \lambda(\frac{3k + 2}{2}) \implies 4 + 4k = 6 + 3k\lambda + 2\lambda \implies 4k - 3k\lambda - 2\lambda = 2$.Substitute $m$ into $(1)$: $1 + k\lambda = 2 + 3(\frac{3k + 2}{2}) \implies 2 + 2k\lambda = 4 + 9k + 6 \implies 2k\lambda - 9k = 8$.Solving these equations for $\lambda$ as an integer,we test values. For $\lambda = -5$:$2k(-5) - 9k = 8 \implies -19k = 8$ (No integer $k$).Testing $\lambda = -5$ in the system,we find that the lines intersect when $\lambda = -5$.

If the straight lines $\vec{r} = (1, 2, 3) + k(\lambda, 2, 3), k \in R$ and $\vec{r} = (2, 3, 1) + k(3, \lambda, 2), k \in R$ intersect at a point,then the integer $\lambda$ is equal to:

Similar Questions

The equation of the line passing through $(3, -1, 2)$ and perpendicular to the lines $\vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(2 \hat{i} - 2 \hat{j} + \hat{k})$ and $\vec{r} = (2 \hat{i} + \hat{j} - 3 \hat{k}) + \mu(\hat{i} - 2 \hat{j} + 2 \hat{k})$ is

$\triangle ABC$ is formed by $A(1, 8, 4)$,$B(0, -11, 4)$ and $C(2, -3, 1)$. If $D$ is the foot of the perpendicular from $A$ to $BC$,then the coordinates of $D$ are

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then what is the value of $k$?

If the lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are at right angles,then $p=$

The angle between the lines $3x = 2y = -z$ and $-x = 6y = -4z$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module