Question

The value of $\lambda$, for which the four points
$\hat{i}-\hat{j}+\hat{k},2\hat{i}+2\hat{j}-3\hat{k},3\hat{i}+2\hat{j}-2\hat{k},\hat{i}+\lambda \hat{j}+2\hat{k}$ are coplanar, is:

• A. $\frac{-8}{5}$
• B. $\frac{-7}{4}$
• C. $\frac{4}{7}$
• D. $\frac{-4}{7}$

Answer 1

Answer: (A)

$\frac{-8}{5}$

Here,

$\overrightarrow{OA}=\overrightarrow{i}-\overrightarrow{j}+\overrightarrow{k}$

$\overrightarrow{OB}=2\overrightarrow{i}+2\overrightarrow{j}-3\overrightarrow{k}$

$\overrightarrow{OC}=3\overrightarrow{i}+2\overrightarrow{j}-2\overrightarrow{k}$

$\overrightarrow{OD}=\overrightarrow{i}+\lambda \overrightarrow{j}+2\overrightarrow{k}$

We know that,

$\text{A, B, C, D are coplanar, if}\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AD}arecoplanar$

and

$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

Solution:

$\overrightarrow{AB}=\overrightarrow{i}+3\overrightarrow{j}-4\overrightarrow{k}$

$\overrightarrow{BC}=2\overrightarrow{i}+3\overrightarrow{j}-3\overrightarrow{k}$

$\overrightarrow{CD}=(\lambda +1)\overrightarrow{j}+\overrightarrow{k}$

Now,

$\Rightarrow \left|\begin{array}{ccc}1& 3& -4\\ 2& 3& -3\\ 0& \lambda +1& 1\end{array}\right|=0$

$\Rightarrow (3-(-3\left)\right(\lambda +1\left)\right)-3\left(2\right)+(-4)\left(2\right(\lambda +1\left)\right)=0$

$\Rightarrow 3+3\lambda +3-6-4(2\lambda +2)=0$

$\Rightarrow 6+3\lambda -6-8\lambda -8=0$

$\Rightarrow 3\lambda -8\lambda -8=0$

$\Rightarrow \lambda =-\frac{8}{5}$