Two sides of a triangle are given by i^+j^+k^ and −i^+2j^+3k^, then area of triangle is

A¯=i^+j^+k^B¯=−i^+2j^+3k^A¯×B¯=ijk111−123=i(3−2)−j(3+1)−k(2+1)=i−4j−3k|A¯×B¯|=12+(+4)2+(−3)2=26∴ Area of Δ4=12|A¯×B¯|=262

Two sides of a triangle are given by i^+j^+k^ and −i^+2j^+3k^, then area of triangle is

A¯=i^+j^+k^B¯=−i^+2j^+3k^A¯×B¯=ijk111−123=i(3−2)−j(3+1)−k(2+1)=i−4j−3k|A¯×B¯|=12+(+4)2+(−3)2=26∴ Area of Δ4=12|A¯×B¯|=262

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Two sides of a triangle are given by $\hat{i} + \hat{j} + \hat{k} and - \hat{i} + 2 \hat{j} + 3 \hat{k}$ , then area of triangle is

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$\sqrt{46}$

$\sqrt{26}$

$\frac{\sqrt{26}}{2}$

answer is B.

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Detailed Solution

$\begin{array}{l} \bar{A} = \hat{i} + \hat{j} + \hat{k} \bar{B} = - \hat{i} + 2 \hat{j} + 3 \hat{k} \\ \bar{A} \times \bar{B} = |\begin{array}{ccc} i & j & k \\ 1 & 1 & 1 \\ - 1 & 2 & 3 \end{array}| = i (3 - 2) - j (3 + 1) - k (2 + 1) \\ = i - 4 j - 3 k \\ | \bar{A} \times \bar{B} | = \sqrt{1^{2} + (+ 4)^{2} + (- 3)^{2}} = \sqrt{26} \\ ∴ Area of Δ^{4} = \frac{1}{2} | \bar{A} \times \bar{B} | = \frac{\sqrt{26}}{2} \end{array}$