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MHT CET · Maths · Vector Algebra

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. A \(\sqrt{71}\) sq. units
  2. B \(\sqrt{29}\) sq. units
  3. C \(\sqrt{41}\) sq. units
  4. D \(\sqrt{61}\) sq. units
Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt{61}\) sq. units

Step-by-step Solution

Detailed explanation

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