Find the reflection of the point (1, 6, 3) in | vedclass

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

Question

(D) Let $P(1, 6, 3)$ be the given point and $L$ be the foot of the perpendicular drawn from $P$ to the given line.The general point on the given line

Vedclass · Accepted Answer

$(1, 0, 7)$

Vedclass · Answer

(D) Let $P(1, 6, 3)$ be the given point and $L$ be the foot of the perpendicular drawn from $P$ to the given line.The general point on the given line is given by $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda$.Thus,the coordinates of $L$ are $(\lambda, 2\lambda + 1, 3\lambda + 2)$.The direction ratios of $PL$ are $(\lambda - 1, 2\lambda + 1 - 6, 3\lambda + 2 - 3)$,which simplifies to $(\lambda - 1, 2\lambda - 5, 3\lambda - 1)$.Since $PL$ is perpendicular to the line with direction ratios $(1, 2, 3)$,we have:$1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$$\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$$14\lambda - 14 = 0 \implies \lambda = 1$.Substituting $\lambda = 1$ into the coordinates of $L$,we get $L(1, 3, 5)$.Let $Q(x_1, y_1, z_1)$ be the reflection of $P(1, 6, 3)$ in the line. Then $L$ is the midpoint of $PQ$.$\frac{x_1 + 1}{2} = 1 \implies x_1 = 1$$\frac{y_1 + 6}{2} = 3 \implies y_1 = 0$$\frac{z_1 + 3}{2} = 5 \implies z_1 = 7$Thus,the reflection of the point $P(1, 6, 3)$ in the given line is $(1, 0, 7)$.

Find the reflection of the point $(1, 6, 3)$ in the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3}$.

Similar Questions

If the orthocentre and the centroid of a triangle are at $(5,2,-6)$ and $(9,6,-4)$ respectively,then its circumcentre is

If $A=(2,3,4)$ and $B=(-2,3,4)$,then the locus of a point $P(x,y,z)$ such that $PA+PB=4$ is

If $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$, then

For $a \neq 0$,if the sum of the distances of a point $P(x, y, z)$ from the points $F_1(a, 0, 0)$ and $F_2(-a, 0, 0)$ is a constant $2k$,then the locus of that point is

Let $A=(1,2,0)$, $B=(2,0,-1)$, $C=(0,-2,3)$ and $D=(-1,2,-3)$ be four points in space. Let $G_1$ be the centroid of triangle $ABC$ and $G_2$ be the centroid of tetrahedron $ABCD$. If $P$ divides $G_1G_2$ in the ratio $4:3$ internally, then $P=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module