If P 0-1-2- Q 4-2-1 and O 0-0-0 are three points- then - POQ -A- 4-6B- n-4C-D- w-2

The correct option is D π/2 PQ^2=(4 0)^2+( 2 1)^2+(1 2)^2 = 16 + 9 + 1 = 26 OP^2= (0 0)^2+(1 0)^2+(2 0)^2 = 0 + 1 + 4 = 5 QO^2= (0 4)^2+(0+2)^2+(0 1)^2 = 16 + ...

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Byju's Answer

Standard VII

Mathematics

Transversal of Parallel Lines

If P 0,1,2, Q...

Question

If P(0, 1, 2), Q(4, -2, 1) and O(0, 0, 0) are three points, then -POQ =

$\frac{π}{6}$

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$\frac{π}{4}$

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$\frac{π}{3}$

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$\frac{π}{2}$

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Solution

The correct option is D
$\frac{π}{2}$

$P Q^{2} = (4 - 0)^{2} + (- 2 - 1)^{2} + (1 - 2)^{2}$

= 16 + 9 + 1 = 26

$O P^{2} = (0 - 0)^{2} + (1 - 0)^{2} + (2 - 0)^{2}$

= 0 + 1 + 4 = 5

$Q O^{2} = (0 - 4)^{2} + (0 + 2)^{2} + (0 - 1)^{2}$

= 16 + 1 + 4 = 21

Since, $P Q^{2} = O P^{2} + Q O^{2}$

Hence, $∠ P O Q = \frac{π}{2}$

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Similar questions

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\frac{π}{6}

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If P(0, 1, 2), Q(4, -2, 1) and O(0, 0, 0) are three points, then -POQ =

The correct option is D π2 PQ2=(4−0)2+(−2−1)2+(1−2)2 = 16 + 9 + 1 = 26 OP2=(0−0)2+(1−0)2+(2−0)2 = 0 + 1 + 4 = 5 QO2=(0−4)2+(0+2)2+(0−1)2 = 16 + 1 + 4 = 21 Since, PQ2=OP2+QO2 Hence, ∠POQ=π2