JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let in a \(\triangle A B C\), the length of the side \(A C\) be 6 , the vertex \(B\) be \((1,2,3)\) and the vertices \(A, C\) lie on the line \(\frac{x-6}{3}=\frac{y-7}{2}=\frac{z-7}{-2}\). Then the area (in sq. units) of \(\triangle \mathrm{ABC}\) is:
- A \(17\)
- B \(21\)
- C \(56\)
- D \(42\)
Answer & Solution
Correct Answer
(B) \(21\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { Let } \mathrm{M}(3 \lambda+6,2 \lambda+7,-2 \lambda+7) \\ & \overrightarrow{\mathrm{BM}}=(3 \lambda+5) \hat{\mathrm{i}}+(2 \lambda+5) \hat{\mathrm{j}}+(-2 \lambda+4) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{AC}} \cdot…
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