JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\mathrm{ABC}\) be a triangle of area \(15 \sqrt{2}\) and the vectors \(\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}, \quad \overrightarrow{B C}=a \hat{i}+b \hat{j}+c \hat{k}\) and \(\overrightarrow{\mathrm{AC}}=6 \hat{\mathrm{i}}+\mathrm{d} \hat{\mathrm{j}}-2 \hat{\mathrm{k}}, \mathrm{d}>0\). Then the square of the length of the largest side of the triangle \(\mathrm{ABC}\) is ...........
- A \(54\)
- B \(45\)
- C \(49\)
- D \(71\)
Answer & Solution
Correct Answer
(A) \(54\)
Step-by-step Solution
Detailed explanation
Area \(=\frac{1}{2}\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 1 & 2 & -7 \\ 6 & \mathrm{~d} & -2\end{array}\right|=15 \sqrt{2}\) \( (-4+7 d) \hat{i}-\hat{j}(-2+42)+\hat{k}(d-12) \) \( (7 d-4)^2+(40)^2+(d-12)^2=1800 \) \( 50 d^2-80 d-40=0 \)…
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