If the lines frac{x-1}{2}=frac{y+2}{3}=frac{z | vedclass

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

Question

(A) The given equations of the lines are $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.The lines pass thro

Vedclass · Accepted Answer

$\frac{7}{2}$

Vedclass · Answer

(A) The given equations of the lines are $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$.The lines pass through points $P_1(1, -2, 1)$ and $P_2(3, k, 0)$ with direction ratios $\vec{v_1} = (2, 3, 4)$ and $\vec{v_2} = (1, 2, 1)$.For two lines to intersect,the shortest distance between them must be zero,which implies the scalar triple product of the vector joining the points and the direction vectors must be zero:$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \ a_1 & b_1 & c_1 \ a_2 & b_2 & c_2 \end{array}ight|=0$Substituting the values:$\left|\begin{array}{ccc} 3-1 & k-(-2) & 0-1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{array}ight|=0$$\left|\begin{array}{ccc} 2 & k+2 & -1 \ 2 & 3 & 4 \ 1 & 2 & 1 \end{array}ight|=0$Expanding the determinant along the first row:$2(3(1) - 4(2)) - (k+2)(2(1) - 4(1)) - 1(2(2) - 3(1)) = 0$$2(3-8) - (k+2)(2-4) - 1(4-3) = 0$$2(-5) - (k+2)(-2) - 1(1) = 0$$-10 + 2k + 4 - 1 = 0$$2k - 7 = 0$$k = \frac{7}{2}$

If the lines $\frac{x-1}{2}=\frac{y+2}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect,then $k$ has the value

Similar Questions

Find the foot of the perpendicular from the point $(2, 3, -8)$ to the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$. Also,find the perpendicular distance from the given point to the line.

If the lines $x = ay + b, z = cy + d$ and $x = a'z + b', y = c'z + d'$ are perpendicular,then

$A$ line drawn from a point $A(-2,-2,3)$ and parallel to the line $\frac{x}{-2}=\frac{y}{2}=\frac{z}{-1}$ meets the $YOZ-$ plane in point $P$. Then,the coordinates of the point $P$ are:

The equation of the line passing through the point of intersection of $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-4}{5}=\frac{y-1}{2}=z$ and also through the point $(2,1,-2)$ is

The angle between the lines $\vec{r}=(2 \hat{i}+\hat{j}-3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})$ and $\frac{x-1}{1}=\frac{y+2}{3}=\frac{z-3}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module