Question

The position vectors of the points \(A, B, C\) and \(D\) are \(3 \hat{i}-2 \hat{j}-\hat{k}, 2 \hat{i}-3 \hat{j}+2 \hat{k}, 5 \hat{i}-\hat{j}+2 \hat{k}\) and \(4 \hat{i}-\hat{j}+\lambda \hat{k}\) respectively. If the points \(A, B, C\) and \(D\) lie on a plane, the value of \(\lambda\) is
(A) 0
(B) 1
(C) 2
(D) \(-4\)

Answer 1

Ans: (D)
Hint : \(\overrightarrow{A B}=-\vec{i}-\vec{j}+3 \vec{k}\)
$$
\left|\begin{array}{ccc}
-1 & -1 & 3 \\
2 & 1 & 3 \\
1 & 1 & \lambda+1
\end{array}\right|=0
$$
$$
\overrightarrow{A C}=2 \vec{i}+\vec{j}+3 \vec{k}
$$
\(\overrightarrow{A D}=\vec{i}+\vec{j}+(\lambda+1) \vec{k}\)
$$
\begin{aligned}
&\left|\begin{array}{ccc}
0 & -1 & 3 \\
1 & 1 & 3 \\
0 & 1 & \lambda+1
\end{array}\right|=0 \\
&\quad-(-\lambda-1-3)=0, \quad(\lambda+4)=0, \lambda=-4
\end{aligned}
$$

Answer 2

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Answer 3

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