The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $\mathrm{X}$-axis at A, Y-axis at C, then the…

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the $\mathrm{X}$-axis at A, Y-axis at C, then the area of $\triangle \mathrm{ABC}=$
  1. $\sqrt{71}$ sq. units
  2. $\sqrt{29}$ sq. units
  3. $\sqrt{41}$ sq. units
  4. $\sqrt{61}$ sq. units

Solution

The vertices of triangle $\mathrm{ABC}$ are $\begin{aligned} & \mathrm{A}=(2,0,0) ; \mathrm{B}=(0,3,0) ; \mathrm{C}=(0,0,4) \\ & \overline{\mathrm{AB}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}} \text { and } \overline{\mathrm{AC}}=-2 \hat{\mathrm{i}}+4 \hat{\mathrm{k}} \\ & \overline{\mathrm{AB}} \times \overline{\mathrm{AC}}=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -2 & 3 & 0 \\ -2 & 0 & 4 \end{array}\right|=\hat{\mathrm{i}}(12)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(6)=12 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+6 \hat{\mathrm{k}} \\ & \therefore \mathrm{A}(\triangle \mathrm{ABC})=\frac{1}{2}|\overline{\mathrm{AB}} \times \overline{\mathrm{AC}}| \\ & =\frac{1}{2}(\sqrt{144+64+36}) \\ & =\sqrt{61} \text { sq. units } \\ & \end{aligned}$

Asked in: MHT CET 2021 (22 Sep Shift 1)

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