Let a → be a vector which is perpendicular to the vector 3 i ^ + 1 2 j ^ + 2 k ^ . If a → * ( 2 i ^ + k ^ ) = 2 i ^ − 1 3 j ^ − 4 k ^ , then the projection of the vector a → on the vector 2 i ^ + 2 j ^ + k ^ is:

a → ⊥ t o 3 i ^ + 1 2 j ^ + 2 k ^ a → * ( 2 i ^ + k ^ ) = 2 i ^ − 1 3 j ^ − 4 k ^ a → ⋅ ( 3 i ^ + 1 2 j ^ + 2 k ^ ) = 0 ( x i + y j ^ + z k ^ ) * ( 2 i ^ + k ^ ) Let a → = ( x i ^ + y j ^ + 2 k ^ ) ( x i ^ + y j ^ + z k ^ ) . ( 3 i ^ + 1 2 j ^ + 2 k ^ ) = 0 = | i j k x y z 2 0 1 | 3 x + y 2 + 2 z = 0 = i ( y − 0 ) − j ( x − 2 z ) + k ( x . 0 − 2 y ) 4x – 12 = 0 y = 2 x = 3 z = -5 x -2z = 13 ∴ a → = ( 3 i ^ + 2 j ^ − 5 k ^ ) a → ⋅ b → | b | = | a c o s θ | b → = 2 i ^ + 2 j ^ + k ^ a → ⋅ i ^ = 6 + 4 − 5 = 5 a → = 3 i ^ + 2 j ^ − 5 k ⇒ ( 5 3 ) = | a c o s θ |

Let $\vec{a}$ be a vector which is perpendicular to the vector $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k} .$ If $\vec{a} * (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k},$ then the projection of the vector $\vec{a}$ on the vector $2 \hat{i} + 2 \hat{j} + \hat{k}$ is:

Option 1 - <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 2 - 1

Option 3 - <math> <mrow> <mfrac> <mrow> <mn>5</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

Option 4 - <math> <mrow> <mfrac> <mrow> <mn>7</mn> </mrow> <mrow> <mn>3</mn> </mrow> </mfrac> </mrow> </math>

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Answered by

Alok Kumar Singh | Contributor-Level 10

6 months ago

Correct Option - 3

Detailed Solution:

$\vec{a} ⊥ t o 3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$ $\vec{a} * (2 \hat{i} + \hat{k}) = 2 \hat{i} - 13 \hat{j} - 4 \hat{k}$

$\vec{a} \cdot (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}) = 0$ $(x i + y \hat{j} + z \hat{k}) * (2 \hat{i} + \hat{k})$

Let $\vec{a} = (x \hat{i} + y \hat{j} + 2 \hat{k})$

$(x \hat{i} + y \hat{j} + z \hat{k}) . (3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}) = 0$ $= | \begin{matrix} i & j & k \\ x & y & z \\ 2 & 0 & 1 \end{matrix} |$