Question

Prove that Ā= (4î – 6ị +8k) and B = (2î + 4ſ + 2k) are mutually perpendicular.

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Answer 1

$\underline{\rm\red{Question }}$

Prove that $\rm \vec{A} = 4\hat{\imath} - 6\hat{\jmath} + 8\hat{k}$ and $\rm \vec{B} = 2\hat{\imath} + 4\hat{\jmath} + 2\hat{k}$ are mutually perpendicular.

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$\underline{\rm\red{Answer }}$

$\underline{\bf\pink{Given - }}$

$\rm \vec{A} = 4\hat{\imath} - 6\hat{\jmath} + 8\hat{k}$

$\rm \vec{B} = 2\hat{\imath} + 4\hat{\jmath} + 2\hat{k}$

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$\underline{\bf\pink{To\: Prove -}}$

$\rm \vec{A}$ and $\rm \vec{B}$ are mutually perpendicular.

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$\underline{\rm\red{Solution - }}$

When the vectors are perpendicular , the dot product of two vectors is equal to zero.

For two vectors -

$\rm \vec A = a_x\hat{\imath} + a_y\hat{\jmath} + a_z\hat{k}$

$\rm \vec B = b_x\hat{\imath} + b_y\hat{\jmath} + b_z\hat{k}$

$\boxed{\rm\purple{ \vec A • \vec B = a_xb_x + a_yb_y + a_zb_z}}$

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For given vectors -

$\rm \vec{A} = 4\hat{\imath} - 6\hat{\jmath} + 8\hat{k}$

$\rm \vec{B} = 2\hat{\imath} + 4\hat{\jmath} + 2\hat{k}$

$\rm \vec A • \vec B = (4 \times 2) + ( - 6 \times 4) + (8 \times 2)$

$\implies$$\rm \vec A • \vec B = 8 - 24 + 16$

$\implies$$\rm \vec A • \vec B = 24 - 24$

$\implies$$\rm \vec A • \vec B = 0$

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As the dot product of the given vectors is equal to zero, both the vectors are mutually perpendicular.

Hence Proved.

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