Vectors 3 Question 30
30. The position vectors of the points $A, B, C$ and $D$ are $3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, 2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}},-\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\lambda \hat{\mathbf{k}}$, respectively. If the points $A, B, C$ and $D$ lie on a plane, find the value of $\lambda$.
$\left(1986,2 \frac{1}{2}\right.$ M)
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Solution:
- Here,
$$ \overrightarrow{\mathbf{A B}}=-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} $$
$\overrightarrow{\mathbf{A C}}=-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{A D}}=\hat{\mathbf{i}}+7 \hat{\mathbf{j}}+(\lambda+1) \hat{\mathbf{k}}$
We know that, $A, B, C, D$ lie in a plane if $\overrightarrow{\mathbf{A B}}, \overrightarrow{\mathbf{A C}}, \overrightarrow{\mathbf{A D}}$ are coplanar i.e.
$$ \begin{array}{ccc} & {[\overrightarrow{\mathbf{A B}} \overrightarrow{\mathbf{A C}} \overrightarrow{\mathbf{A D}}]=0 \Rightarrow\left|\begin{array}{ccc} -1 & 5 & -3 \\ -4 & 3 & 3 \\ 1 & 7 & \lambda+1 \end{array}\right|=0} \\ \Rightarrow & -1(3 \lambda+3-21)-5(-4 \lambda-4-3)-3(-28-3)=0 \\ \Rightarrow & -17 \lambda+146=0 \\ \therefore & \lambda=\frac{146}{17} \end{array} $$
