If $\mathrm{A}(3,2,-1), \mathrm{B}(-2,2,-3)$ and $\mathrm{D}(-2,5,-4)$ are the vertices of a parallelogram,…

If $\mathrm{A}(3,2,-1), \mathrm{B}(-2,2,-3)$ and $\mathrm{D}(-2,5,-4)$ are the vertices of a parallelogram, then the area of the parallelogram is
  1. 296 sq. units
  2. $\sqrt{286}$ sq. units
  3. 300 sq. units
  4. $\sqrt{300}$ sq. units

Solution

We have $\mathrm{A} \equiv(3,2,-1), \mathrm{B} \equiv(-2,2,-3)$ and $\mathrm{D} \equiv(-2,5,-4)$ $\therefore \overline{\mathrm{AB}}=-5 \hat{\mathrm{i}}-2 \hat{\mathrm{k}} \text { and } \overline{\mathrm{AD}}=-5 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-3 \hat{\mathrm{k}}$ Area of parallelogram $=|\overline{\mathrm{AB}} \times \overline{\mathrm{AD}}|$ Now $\begin{aligned} & \overline{\mathrm{AB}} \times \overline{\mathrm{AD}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -5 & 0 & -2 \\ -5 & 3 & -3 \end{array}\right|=\hat{\mathrm{i}}(6)-\hat{\mathrm{j}}(5)+\hat{\mathrm{k}}(-15)=6 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\ & \therefore \text { Area }=\sqrt{(6)^2+(-5)^2+(-15)^2} \end{aligned}$

Asked in: MHT CET 2021 (23 Sep Shift 2)

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