Question

If

A (3, 2, - 1), B (- 2, 2, - 3), C (3, 5, - 2) & D (- 2, 5, - 4)

then (i) verify that the points are the vertices of a parallelogram and (ii) find its area.

Accepted Answer

Given,

$A (3, 2, - 1), B (- 2, 2, - 3), C (3, 5, - 2) & D (- 2, 5, - 4)$ ,

Now let $\vec{a}, \vec{b}, \vec{c} & \vec{d}$ be position vector of $A, B, C & D$ respectively with respect to origin,

So, $\vec{a} = 3 \hat{i} + 2 \hat{j} - \hat{k}, \vec{b} = - 2 \hat{i} + 2 \hat{j} - 3 \hat{k}, \vec{c} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k} & \vec{d} = - 2 \hat{i} + 5 \hat{j} - 4 \vec{k}$

Now $\vec{A B} = \vec{b} - \vec{a} = - 2 \hat{i} + 2 \hat{j} - 3 \hat{k} - (3 \hat{i} + 2 \hat{j} - \hat{k}) = - 5 \hat{i} - 2 \hat{k}$ $. . . . . . . . . (1)$

And $\vec{D C} = \vec{c} - \vec{d} = 3 \hat{i} + 5 \hat{j} - 2 \hat{k} - (- 2 \hat{i} + 5 \hat{j} - 4 \vec{k}) = 5 \hat{i} + 2 \vec{k} = - (- 5 \hat{i} - 2 \vec{k})$ $. . . . . . (2)$

Now from both the equation we can say that $\vec{A B} = λ \vec{D C} \{where λ = - 1\}$ or $\vec{A B}$ is parallel to $\vec{D C}$ ,

Also $|\vec{A B}| = |\vec{D C}| = \sqrt{5^{2} + 2^{2}} = \sqrt{29}$

So we can say that $A B C D$ is a parallelogram,

Now area will be $|\vec{A B} \times \vec{A D}| = |(- 5 \hat{i} - 2 \hat{k}) \times (- 5 \hat{i} + 3 \hat{j} - 3 \hat{k})|$

$\Rightarrow |\vec{A B} \times \vec{A D}| = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ - 5 & 0 & - 2 \\ - 5 & 3 & - 3 \end{matrix}| = |6 \hat{i} - 5 \hat{j} - 15 \hat{k}| = \sqrt{6^{2} + 5^{2} + 15^{2}} = \sqrt{286}$ square units.

Maharashtra Board Solutions for Chapter: Vectors, Exercise 6: Miscellaneous Exercise 5

Maharashtra Board Mathematics Solutions for Exercise - Maharashtra Board Solutions for Chapter: Vectors, Exercise 6: Miscellaneous Exercise 5

Questions from Maharashtra Board Solutions for Chapter: Vectors, Exercise 6: Miscellaneous Exercise 5 with Hints & Solutions

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