Find the coordinates of the foot of the perpe | vedclass

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

Question

(A) Let the line passing through $B(4, 7, 1)$ and $C(3, 5, 3)$ be $L$. The direction ratios of $BC$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation o

Vedclass · Accepted Answer

$\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)$

Vedclass · Answer

(A) Let the line passing through $B(4, 7, 1)$ and $C(3, 5, 3)$ be $L$. The direction ratios of $BC$ are $(3-4, 5-7, 3-1) = (-1, -2, 2)$.The equation of the line $BC$ is $\frac{x-4}{-1} = \frac{y-7}{-2} = \frac{z-1}{2} = k$.Any point $D$ on the line $BC$ is given by $(4-k, 7-2k, 1+2k)$.Since $AD \perp BC$,the direction ratios of $AD$ are $(4-k-1, 7-2k-0, 1+2k-3) = (3-k, 7-2k, 2k-2)$.The dot product of the direction ratios of $AD$ and $BC$ must be zero:$-1(3-k) - 2(7-2k) + 2(2k-2) = 0$.$-3 + k - 14 + 4k + 4k - 4 = 0$.$9k - 21 = 0 \Rightarrow k = \frac{21}{9} = \frac{7}{3}$.Substituting $k = \frac{7}{3}$ into the coordinates of $D$:$x = 4 - \frac{7}{3} = \frac{5}{3}$,$y = 7 - 2(\frac{7}{3}) = 7 - \frac{14}{3} = \frac{7}{3}$,$z = 1 + 2(\frac{7}{3}) = 1 + \frac{14}{3} = \frac{17}{3}$.Thus,the coordinates of the foot of the perpendicular are $\left( \frac{5}{3}, \frac{7}{3}, \frac{17}{3} \right)$.

Find the coordinates of the foot of the perpendicular drawn from the point $A(1, 0, 3)$ to the line joining the points $B(4, 7, 1)$ and $C(3, 5, 3)$.

Similar Questions

$ABC$ is a triangle in a plane with vertices $A(2, 3, 5)$,$B(-1, 3, 2)$,and $C(\lambda, 5, \mu)$. If the median through $A$ is equally inclined to the coordinate axes,then the value of $\lambda + \mu$ is:

The coordinates of the points on the line $\frac{x+2}{1}=\frac{y-1}{2}=\frac{z+1}{-2}$ at a distance of $12 \text{ units}$ from the point $A(-2, 1, -1)$ are

The Cartesian equation of the line which is parallel to the line $\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$ and passing through the point $(1, -3, 5)$ is:

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

If the Cartesian equation of a line is $6x-2=3y+1=2z-2$,then the vector equation of the line is

Vedclass Test Series

Exam Paper Generator

Online Exam Module