Find a vector x→ which is perpendicular to both A→ and B→ but has magnitude equal to that ofB→ . Rule: Inter change coeff. Of i^ and j^ and change sign of one of the vectors, A→= 3i^−2j^+k^,B→=4i^+3j^−2k^

B→=42+32+(-2)2=29A→×B→= |i^j^k^3−2143−2| =i^[(4-3)-j^-6-4+k^9+8A→×B→=12+102+172=390n→= A→×B→|A→×B→|= i^+10j^+17k^390x→= |B→|n^= 29(i^+10j^+17k^)390

Find a vector x→ which is perpendicular to both A→ and B→ but has magnitude equal to that ofB→ . Rule: Inter change coeff. Of i^ and j^ and change sign of one of the vectors, A→= 3i^−2j^+k^,B→=4i^+3j^−2k^

B→=42+32+(-2)2=29A→×B→= |i^j^k^3−2143−2| =i^[(4-3)-j^-6-4+k^9+8A→×B→=12+102+172=390n→= A→×B→|A→×B→|= i^+10j^+17k^390x→= |B→|n^= 29(i^+10j^+17k^)390

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Find a vector $\vec{x}$ which is perpendicular to both $\vec{A}$ and $\vec{B}$ but has magnitude equal to that of $\vec{B}$ .
Rule: Inter change coeff. Of $\hat{i}$ and $\hat{j}$ and change sign of one of the vectors,

$\vec{A} = 3 \hat{i} - 2 \hat{j} + \hat{k}, \vec{B} = 4 \hat{i} + 3 \hat{j} - 2 \hat{k}$

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$\frac{1}{\sqrt{10}} (\hat{i} + 10 \hat{j} + 17 \hat{k})$

$\frac{1}{\sqrt{10}}$ $(\hat{i} - 10 \hat{j} + 17 \hat{k})$

$\sqrt{\frac{29}{390}} (\hat{i} - 10 \hat{j} + 17 \hat{k})$

$\sqrt{\frac{29}{390}} (\hat{i} + 10 \hat{j} + 17 \hat{k})$

answer is D.

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Detailed Solution

$|\vec{B}| = \sqrt{4^{2} + 3^{2} + {(- 2)}^{2}} = \sqrt{29}$

$\vec{A} \times \vec{B} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & - 2 & 1 \\ 4 & 3 & - 2 \end{matrix} |$ $= \hat{i} [(4 - 3) - \hat{j} (- 6 - 4) + \hat{k} (9 + 8)$

$|\vec{A} \times \vec{B}| = \sqrt{1^{2} + 10^{2} + 17^{2}} = \sqrt{390}$

$\vec{n} = \frac{\vec{A} \times \vec{B}}{| \vec{A} \times \vec{B} |} = \frac{\hat{i} + 10 \hat{j} + 17 \hat{k}}{\sqrt{390}}$

$\vec{x} = | \vec{B} | \hat{n} = \frac{\sqrt{29} (\hat{i} + 10 \hat{j} + 17 \hat{k})}{\sqrt{390}}$

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Find a vector x→ which is perpendicular to both A→ and B→ but has magnitude equal to that ofB→ . Rule: Inter change coeff. Of i^ and j^ and change sign of one of the vectors, A→= 3i^−2j^+k^,B→=4i^+3j^−2k^

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