The coordinates of the foot of the perpendicu | vedclass

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

Question

(A) The given line is $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$.Rewriting it in standard form: $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac

Vedclass · Accepted Answer

$(3, 5, 9)$

Vedclass · Answer

(A) The given line is $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$.Rewriting it in standard form: $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} = k$.Any point $M$ on the line is given by $(3k + 6, 2k + 7, -2k + 7)$.Let $A = (1, 2, 3)$. The vector $\vec{AM} = (3k + 6 - 1, 2k + 7 - 2, -2k + 7 - 3) = (3k + 5, 2k + 5, -2k + 4)$.The direction vector of the line is $\vec{v} = (3, 2, -2)$.Since $\vec{AM} \perp \vec{v}$,their dot product is zero: $\vec{AM} \cdot \vec{v} = 0$.$3(3k + 5) + 2(2k + 5) - 2(-2k + 4) = 0$.$9k + 15 + 4k + 10 + 4k - 8 = 0$.$17k + 17 = 0 \Rightarrow k = -1$.Substituting $k = -1$ into the coordinates of $M$: $x = 3(-1) + 6 = 3$,$y = 2(-1) + 7 = 5$,$z = -2(-1) + 7 = 9$.Thus,the coordinates of the foot of the perpendicular are $(3, 5, 9)$.

The coordinates of the foot of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ are:

Similar Questions

Show that the line joining the origin to the point $(2,1,1)$ is perpendicular to the line determined by the points $(3,5,-1)$ and $(4,3,-1).$

If the shortest distance between the lines $\frac{x-k}{2}=\frac{y-4}{3}=\frac{z-3}{4}$ and $\frac{x-2}{4}=\frac{y-4}{6}=\frac{z-7}{8}$ is $\frac{13}{\sqrt{29}}$,then $k=$

Let $a, b \in R$. If the mirror image of the point $P(a, 6, 9)$ with respect to the line $\frac{x-3}{7} = \frac{y-2}{5} = \frac{z-1}{-9}$ is $(20, b, -a-9)$,then $|a+b|$ is equal to

The length and foot of the perpendicular from the point $(2, -1, 5)$ to the line $\frac{x - 11}{10} = \frac{y + 2}{-4} = \frac{z + 8}{-11}$ are

Show that the points $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$ are collinear.

Vedclass Test Series

Exam Paper Generator

Online Exam Module