Given two vectors a→=−i^+j^+2k^ and b→=−i^−2j^−k^ (a-s; b-r; c-q; d-p)

(a_a→×b→=i^j^k^−112−1−2−1=3i^−3j^+3k^Hence, the area of the triangle is 332b) The area of the paralllelogram is 33c) The area of a parallelogram whose diagonals are 2a→ and 4b→ is 12|2a→×4b→|=123d) Volume of the parallelepiped =|(a→×b→)⋅c→|=3−3+3=3

Given two vectors a→=−i^+j^+2k^ and b→=−i^−2j^−k^ (a-s; b-r; c-q; d-p)

(a_a→×b→=i^j^k^−112−1−2−1=3i^−3j^+3k^Hence, the area of the triangle is 332b) The area of the paralllelogram is 33c) The area of a parallelogram whose diagonals are 2a→ and 4b→ is 12|2a→×4b→|=123d) Volume of the parallelepiped =|(a→×b→)⋅c→|=3−3+3=3

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Given two vectors $\vec{a} = - \hat{i} + \hat{j} + 2 \hat{k} and \vec{b} = - \hat{i} - 2 \hat{j} - \hat{k}$ (a-s; b-r; c-q; d-p)

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Area of triangle formed by $\vec{a}$ and $\vec{b}$

Area of parallelogram having sides $\vec{a}$ and $\vec{b}$

Area of parallelogram having diagonals $2 \vec{a}$ and $4 \vec{b}$

Volume of parallelepiped formed by $\vec{a}, \vec{b}$ and $\vec{c} = \hat{i} + \hat{j} + \hat{k}$

$3$

$12 \sqrt{3}$

$3 \sqrt{3}$

$\frac{3 \sqrt{3}}{2}$

answer is , , , A.

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Detailed Solution

(a_ $\vec{a} \times \vec{b} = |\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 1 & 2 \\ - 1 & - 2 & - 1 \end{array}| = 3 \hat{i} - 3 \hat{j} + 3 \hat{k}$