JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec{a}, \vec{b}\) and \(\vec{c}\) be three non-zero non-coplanar vectors. Let the position vectors of four points \(A, B, \quad C\) and \(D\) be \(\vec{a}-\vec{b}+\vec{c}, \lambda \vec{a}-3 \vec{b}+4 \vec{c}\), \(-\vec{a}+2 \vec{b}-3 \vec{c}\) and \(2 \vec{a}-4 \vec{b}+6 \vec{c}\) respectively. If \(\overrightarrow{A B}\), \(\overline{ AC }\) and \(\overline{ AD }\) are coplanar, then \(\lambda\) is :
- A \(4\)
- B \(6\)
- C \(2\)
- D \(8\)
Answer & Solution
Correct Answer
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\overline{A B}=(\lambda-1) \bar{a}-2 \bar{b}+3 \bar{c}\) \(\overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c}\) \(\overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c}\) \(\left|\begin{array}{ccc}\lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5\end{array}\right|=0\)…
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