The Vector OA -2 -2 k- is rotated through an angle 90- about origin OIn new Position it passes through -k- Then the vector in new position isA- 2 -2 -2 k-B- 2 -2 k-C- 32 -2 -k-D- 3-k-2

The correct option is C 3(2î+2ĵ+k̂) Vector perpendicular to a⃗ and lie on the plane of a⃗& b⃗ is parallel to (a⃗×b⃗)×a⃗ =(a⃗.a⃗)b⃗ (b⃗.a⃗)=9(j+k) 3( 2i+j+ ...

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Byju's Answer

Standard XII

Physics

Vectors and Its Types

The Vector OA...

Question

The Vector $- - \to O A = - 2^i +^j + 2^k$ is rotated through an angle $90^{\circ}$ about origin (O)In new Position it passes through $^i +^k$ . Then the vector in new position is

$2^i +^j + 2^k$

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$2^i + 2^j + 2^k$

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$3 (2^i + 2^j +^k)$

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$\frac{3 (^i +^k)}{\sqrt{2}}$

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Solution

The correct option is C
$3 (2^i + 2^j +^k)$

Vector perpendicular to $\to a$ and lie on the plane of $\to a & \to b$ is parallel to $(\to a \times \to b) \times \to a = (\to a . \to a) \to b - (\to b . \to a) = 9 (j + k) - 3 (- 2 i + j + 2 k) = 3 (2 i + 2 j + k)$

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The Vector −−→OA=−2^i+^j+2^k is rotated through an angle 90∘ about origin (O)In new Position it passes through ^i+^k. Then the vector in new position is

The correct option is C 3(2^i+2^j+^k) Vector perpendicular to →a and lie on the plane of →a&→b is parallel to (→a×→b)×→a=(→a.→a)→b−(→b.→a)=9(j+k)−3(−2i+j+2k)=3(2i+2j+k)

The Vector $- - \to O A = - 2^i +^j + 2^k$ is rotated through an angle $90^{\circ}$ about origin (O)In new Position it passes through $^i +^k$ . Then the vector in new position is

The correct option is C
$3 (2^i + 2^j +^k)$

Vector perpendicular to $\to a$ and lie on the plane of $\to a & \to b$ is parallel to $(\to a \times \to b) \times \to a = (\to a . \to a) \to b - (\to b . \to a) = 9 (j + k) - 3 (- 2 i + j + 2 k) = 3 (2 i + 2 j + k)$