Aspire Faculty ID #17299 · Topic: CUET UG 2022 Applied Mathematics · Just now
CUET UG 2022 Applied Mathematics

The area of the parallelogram whose adjacent sides are $(\mathbf{i}+\mathbf{k})$ and $(2\mathbf{i}+\mathbf{j}+\mathbf{k})$ is:

Solution

Area of parallelogram $= |\vec a \times \vec b|$ Let $\vec a = (1,0,1)$ $\vec b = (2,1,1)$ Cross product $\vec a \times \vec b = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix}$ $= \mathbf{i}(0\cdot1-1\cdot1) -\mathbf{j}(1\cdot1-1\cdot2) +\mathbf{k}(1\cdot1-0\cdot2)$ $= -\mathbf{i}+\mathbf{j}+\mathbf{k}$ Magnitude $|\vec a \times \vec b| =\sqrt{(-1)^2+1^2+1^2}$ $=\sqrt{3}$

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