The base of the pyramid AOBC is an equilateral triangle OBC with each side equal to 4 -2- where O is the origin of reference- AO is perpendicular to the plane of OBC and -AO-2 - Then the cosine of the angle between the skew straight lines- one passing through A and the mid-point of OB and the other passing through O and the mid-point of BC- is

Given, |b|=|c|=|b c|=4 √(2) nbsp; |a|=2 a·b=a·c=0 M A=a b2=2 a b2 OD=b+c2 ∴cosθ=MA·OD |MA||OD| ⇒cosθ=(2 a b) ·(b+c)|2 a b||b+c| Numerator =2 a·b 2 a·c b^2 b ...

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Byju's Answer

Standard XIII

Mathematics

Angle between Two Lines

The base of t...

Question

The base of the pyramid $A O B C$ is an equilateral triangle $O B C$ with each side equal to $4 \sqrt{2},$ where $O$ is the origin of reference, $A O$ is perpendicular to the plane of $△ O B C$ and $| - - \to A O | = 2 .$ Then the cosine of the angle between the skew straight lines, one passing through $A$ and the mid-point of $O B$ and the other passing through $O$ and the mid-point of $B C,$ is

- \frac{1}{\sqrt{2}}

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\sqrt{\frac{3}{5}}

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- \sqrt{\frac{3}{5}}

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\frac{1}{\sqrt{2}}

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Solution

The correct option is D $\frac{1}{\sqrt{2}}$

Given, $| \to b | = | \to c | = | \to b - \to c | = 4 \sqrt{2}$
$| \to a | = 2$
$\to a \cdot \to b = \to a \cdot \to c = 0$
$- - \to M A = \to a - \frac{\to b}{2} = \frac{2 \to a - \to b}{2}$
$- - \to O D = \frac{\to b + \to c}{2}$
$∴ cos θ = \frac{- - \to M A \cdot - - \to O D}{| - - \to M A | | - - \to O D |}$
$\Rightarrow cos θ = \frac{(2 \to a - \to b) \cdot (\to b + \to c)}{| 2 \to a - \to b | | \to b + \to c |}$

Numerator $= 2 \to a \cdot \to b - 2 \to a \cdot \to c - {\to b}^{2} - \to b \cdot \to c$ $= 0 + 0 - (32 + 32 \cdot \frac{1}{2}) = - 48$

Denominator : $| 2 \to a - \to b |^{2} = 4 {\to a}^{2} + {\to b}^{2} - 4 \to a \cdot \to b$
$= 16 + 32 = 48$
$\Rightarrow | 2 \to a - \to b | = 4 \sqrt{3}$
and $| \to b + \to c |^{2} = {\to b}^{2} + {\to c}^{2} + 2 \to b \cdot \to c$ $= 32 + 32 + 2 \cdot 32 \cdot \frac{1}{2} = 96$ $∴ | \to b + \to c | = 4 \sqrt{6}$
$\Rightarrow$ Denominator $= 4 \sqrt{6} \cdot 4 \sqrt{3}$ $= 48 \sqrt{2}$

$∴ cos θ = \frac{- 48}{48 \sqrt{2}} = - \frac{1}{\sqrt{2}}$
Hence, $cos θ = \frac{1}{\sqrt{2}}$ or $- \frac{1}{\sqrt{2}}$

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