The two adjacent sides of a parallelogram are | vedclass

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(A) Let the adjacent sides be $\vec{a} = 2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b} = \hat{i}-2 \hat{j}-3 \hat{k}$.The diagonal of the parallelogram

Vedclass · Accepted Answer

Unit vector: $\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$,Area: $11 \sqrt{5}$ sq. units

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(A) Let the adjacent sides be $\vec{a} = 2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\vec{b} = \hat{i}-2 \hat{j}-3 \hat{k}$.The diagonal of the parallelogram is given by $\vec{d} = \vec{a} + \vec{b} = (2+1) \hat{i} + (-4-2) \hat{j} + (5-3) \hat{k} = 3 \hat{i} - 6 \hat{j} + 2 \hat{k}$.The magnitude of the diagonal is $|\vec{d}| = \sqrt{3^2 + (-6)^2 + 2^2} = \sqrt{9 + 36 + 4} = \sqrt{49} = 7$.The unit vector parallel to the diagonal is $\frac{\vec{d}}{|\vec{d}|} = \frac{3 \hat{i} - 6 \hat{j} + 2 \hat{k}}{7} = \frac{3}{7} \hat{i} - \frac{6}{7} \hat{j} + \frac{2}{7} \hat{k}$.The area of the parallelogram is given by $|\vec{a} 	imes \vec{b}|$.$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & -4 & 5 \ 1 & -2 & -3 \end{vmatrix} = \hat{i}(12 - (-10)) - \hat{j}(-6 - 5) + \hat{k}(-4 - (-4)) = 22 \hat{i} + 11 \hat{j} + 0 \hat{k}$.Area $= |22 \hat{i} + 11 \hat{j}| = \sqrt{22^2 + 11^2} = \sqrt{484 + 121} = \sqrt{605} = 11 \sqrt{5}$ square units.

The two adjacent sides of a parallelogram are $2 \hat{i}-4 \hat{j}+5 \hat{k}$ and $\hat{i}-2 \hat{j}-3 \hat{k}.$ Find the unit vector parallel to its diagonal. Also,find its area.

Similar Questions

The unit vector perpendicular to the vector $\hat{i}-2 \hat{j}+3 \hat{k}$ and coplanar with the vectors $\hat{i}+\hat{j}+\hat{k}$ and $2 \hat{i}-\hat{j}-\hat{k}$ is

If $\overrightarrow{a}=\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{b}=\hat{i}+\hat{j}$, $\overrightarrow{c}=\hat{i}$ and $(\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c}=\lambda \overrightarrow{a}+\mu \overrightarrow{b}$, then $\lambda+\mu$ is equal to:

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