Let O be the origin- Let OP-x-y-k- and OQ- -2-3xk- x-y-x-0- besuch that -PQ-20 and the vector OP is perpendicular to OQ- If OR-3-z-7k- z- is coplanar with OP and OQ- then the value of x2-y2-z2 is equal to -

Given OP=xî+yĵ k̂, OQ= nbsp; î+2ĵ+3xk̂ ∵OP⊥OQ ⇒ nbsp; OP·OQ=0 ⇒ x+2y 3x=0 ⇒ nbsp; y=2x⋯(i) Now PQ=( 1 x)î+(2 y)ĵ+(3x+1)k̂ ⇒|PQ|=√(( 1 x)^2+(2 y)^2+(3x+1) ...

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Byju's Answer

Standard XII

Mathematics

Angle between Two Planes

Let O be the ...

Question

Let $O$ be the origin. Let $- - \to O P = x^i + y^j -^k$ and $- - \to O Q = -^i + 2^j + 3 x^k, x, y \in R, x > 0,$ be
such that $∣ ∣ ∣ - - \to P Q ∣ ∣ ∣ = \sqrt{20}$ and the vector $- - \to O P$ is perpendicular to $- - \to O Q$ . If $- - \to O R = 3^i + z^j - 7^k, z \in R$ is coplanar with $- - \to O P$ and $- - \to O Q,$ then the value of $x^{2} + y^{2} + z^{2}$ is equal to :

2

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Solution

The correct option is B $9$
Given $- - \to O P = x^i + y^j -^k, - - \to O Q = -^i + 2^j + 3 x^k$
$∵ - - \to O P ⊥ - - \to O Q$
$\Rightarrow - - \to O P \cdot - - \to O Q = 0$
$\Rightarrow - x + 2 y - 3 x = 0$
$\Rightarrow y = 2 x \dots (i)$

Now $- - \to P Q = (- 1 - x)^i + (2 - y)^j + (3 x + 1)^k$
$\Rightarrow ∣ ∣ ∣ - - \to P Q ∣ ∣ ∣ = \sqrt{{(- 1 - x)}^{2} + {(2 - y)}^{2} + {(3 x + 1)}^{2}} = \sqrt{20}$
$\Rightarrow 20 = 1 + x^{2} + 2 x + 4 + y^{2} - 4 y + 9 x^{2} + 1 + 6 x$
$\Rightarrow 20 = 10 x^{2} + y^{2} + 8 x + 6 - 4 y$
$\Rightarrow 20 = 10 x^{2} + 4 x^{2} + 8 x + 6 - 8 x (∵ y = 2 x)$
$14 = 14 x^{2}$
$\Rightarrow x^{2} = 1 \Rightarrow x = 1 (∵ x > 0)$
and $y = 2 ($ From $(i))$
Since vectors $- - \to O P, - - \to O Q, - - \to O R$ are coplanar
$∴ ∣ ∣ ∣ ∣ \begin{matrix} x & y & - 1 - 1 & 2 & 3 x 3 & z & - 7 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & 2 & - 1 - 1 & 2 & 3 3 & z & - 7 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow 1 (- 14 - 3 z) - 2 (7 - 9) - 1 (- z - 6) = 0$
$\Rightarrow - 14 - 3 z + 4 + z + 6 = 0$
$\Rightarrow 2 z = - 4 \Rightarrow z = - 2$
$∴ x^{2} + y^{2} + z^{2} = 9$

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Similar questions

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