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If a⃗=2j^−k^\vec{a} = 2\hat{j} - \hat{k}, b⃗=2i^−3j^+k^\vec{b} = 2\hat{i} - 3\hat{j} + \hat{k} and c⃗=−i^+k^\vec{c} = -\hat{i} + \hat{k} are three vectors, then the area (in sq. units) of the parallelogram whose diagonals are (b⃗+c⃗)(\vec{b} + \vec{c}) and (a⃗+c⃗)(\vec{a} + \vec{c}) is

Solution

✅ Correct Option: 2

The area of a parallelogram whose diagonals are d1⃗\vec{d_1} and d2⃗\vec{d_2} is given by:

Area=12∣d1⃗×d2⃗∣\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|

  1. Find the diagonals:

d1⃗=b⃗+c⃗=(2i^−3j^+k^)+(−i^+k^)=i^−3j^+2k^\vec{d_1} = \vec{b} + \vec{c} = (2\hat{i} - 3\hat{j} + \hat{k}) + (-\hat{i} + \hat{k}) = \hat{i} - 3\hat{j} + 2\hat{k}

d2⃗=a⃗+c⃗=(2j^−k^)+(−i^+k^)=−i^+2j^\vec{d_2} = \vec{a} + \vec{c} = (2\hat{j} - \hat{k}) + (-\hat{i} + \hat{k}) = -\hat{i} + 2\hat{j}

  1. Compute the cross product:

d1⃗×d2⃗=∣i^j^k^1−32−120∣=−4i^−2j^−k^\vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ -1 & 2 & 0 \end{vmatrix} = -4\hat{i} - 2\hat{j} - \hat{k}

  1. Calculate the area:

∣d1⃗×d2⃗∣=(−4)2+(−2)2+(−1)2=21|\vec{d_1} \times \vec{d_2}| = \sqrt{(-4)^2 + (-2)^2 + (-1)^2} = \sqrt{21}

Area=212 sq. units\text{Area} = \frac{\sqrt{21}}{2} \text{ sq. units}

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