Let A, B, C be three points on overline{OX}, | vedclass

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

Question

(C) Let the coordinates of point $P$ be $(u, v, w)$.Since $P$ is equidistant from $O(0, 0, 0), A(3, 0, 0), B(0, 6, 0),$ and $C(0, 0, 9)$,we have $PO^2

Vedclass · Accepted Answer

$\left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$

Vedclass · Answer

(C) Let the coordinates of point $P$ be $(u, v, w)$.Since $P$ is equidistant from $O(0, 0, 0), A(3, 0, 0), B(0, 6, 0),$ and $C(0, 0, 9)$,we have $PO^2 = PA^2 = PB^2 = PC^2$.$PO^2 = u^2 + v^2 + w^2$.$PA^2 = (u-3)^2 + v^2 + w^2 = u^2 - 6u + 9 + v^2 + w^2$.Equating $PO^2 = PA^2$,we get $u^2 = u^2 - 6u + 9$ $\Rightarrow 6u = 9$ $\Rightarrow u = \frac{3}{2}$.$PB^2 = u^2 + (v-6)^2 + w^2 = u^2 + v^2 - 12v + 36 + w^2$.Equating $PO^2 = PB^2$,we get $v^2 = v^2 - 12v + 36$ $\Rightarrow 12v = 36$ $\Rightarrow v = 3$.$PC^2 = u^2 + v^2 + (w-9)^2 = u^2 + v^2 + w^2 - 18w + 81$.Equating $PO^2 = PC^2$,we get $w^2 = w^2 - 18w + 81$ $\Rightarrow 18w = 81$ $\Rightarrow w = \frac{81}{18} = \frac{9}{2}$.Thus,$P = \left(\frac{3}{2}, 3, \frac{9}{2}\right)$.Given $Q = (2, 5, 8)$,point $R$ divides $PQ$ in the ratio $3:2$. Using the section formula $\left(\frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n}, \frac{mz_2 + nz_1}{m+n}\right)$:$R = \left(\frac{3(2) + 2(\frac{3}{2})}{3+2}, \frac{3(5) + 2(3)}{3+2}, \frac{3(8) + 2(\frac{9}{2})}{3+2}\right)$$R = \left(\frac{6+3}{5}, \frac{15+6}{5}, \frac{24+9}{5}\right) = \left(\frac{9}{5}, \frac{21}{5}, \frac{33}{5}\right)$.

Similar Questions

$A$ line in the $3-$dimensional space makes an angle $\theta \left( 0 < \theta \le \frac{\pi}{2} \right)$ with both the $x$ and $y$ axes. Then the set of all values of $\theta$ is the interval

The projections of a line on the coordinate axes are $2, 3, 6$. Then the length of the line is

Find the angle between the lines whose direction ratios are $1, 1, 2$ and $\sqrt{3}-1, -\sqrt{3}-1, 4$. (in $^{\circ}$)

Let $O$ be the origin and $P$ be a point which is at a distance of $3$ units from the origin. If the direction ratios of $\vec{OP}$ are $(1, -2, -2)$,then the coordinates of $P$ are

The direction cosines of a line are $\langle \frac{-9}{11}, \frac{6}{11}, \frac{-2}{11} \rangle$ respectively. Then its direction ratios are

Vedclass Test Series

Exam Paper Generator

Online Exam Module