The magnitude of the projection of the vector 2 i ^ + 3 j ^ + k ^ on the vector perpendicular to the plane containing the vectors i ^ + j ^ + k ^ and i ^ + 2 j ^ + 3 k ^ , is:

Vector perpendicular to i ^ + j ^ + k ^ and i ^ + 2 j ^ + 3 k ^ = i ^ j ^ k ^ 1 1 1 1 2 3 = i ^ - 2 j ^ + k ^ ∴ Required projection = | ( 2 i ^ + 3 j ^ + k ^ ) ⋅ ( i ^ - 2 j ^ + k ^ ) | | i ^ - 2 j ^ + k ^ | = | 2 - 6 + 1 | 6 = 3 6 = 3 2

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MathematicsVectorDot Product & Its Application ( Projection etc.)Medium2 minPYQ_2019

MathematicsMediumsingle choice

The magnitude of the projection of the vector $2 \hat{i} + 3 \hat{j} + \hat{k}$ on the vector perpendicular to the plane containing the vectors $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k},$ is:

Options:

Answer:

Solution:

Vector perpendicular to $\hat{i} + \hat{j} + \hat{k}$ and $\hat{i} + 2 \hat{j} + 3 \hat{k}$

$= |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & 2 & 3 \end{matrix}| = \hat{i} - 2 \hat{j} + \hat{k}$

$∴$ Required projection

$= \frac{| (2 \hat{i} + 3 \hat{j} + \hat{k}) \cdot (\hat{i} - 2 \hat{j} + \hat{k}) |}{| \hat{i} - 2 \hat{j} + \hat{k} |}$

$= \frac{| 2 - 6 + 1 |}{\sqrt{6}} = \frac{3}{\sqrt{6}} = \sqrt{\frac{3}{2}}$

Stream:JEESubject:MathematicsTopic:VectorSubtopic:Dot Product & Its Application ( Projection etc.)

⏱ 2mℹ️ Source: PYQ_2019

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