Question

Using vectors, find the value of λ such that the points (λ, −10, 3), (1, −1, 3) and (3, 5, 3) are collinear. [NCERT EXEMPLAR]

Answer 1

Let the given points be A(λ, −10, 3), B(1, −1, 3) and C(3, 5, 3).

$\overrightarrow{\mathrm{AB}}=\left(\hat{i}-\hat{j}+3\hat{k}\right)-\left(\lambda \hat{i}-10\hat{j}+3\hat{k}\right)=\left(1-\lambda \right)\hat{i}+9\hat{j}$

$\overrightarrow{\mathrm{AC}}=\left(3\hat{i}+5\hat{j}+3\hat{k}\right)-\left(\lambda \hat{i}-10\hat{j}+3\hat{k}\right)=\left(3-\lambda \right)\hat{i}+15\hat{j}$

If the points A, B, C are collinear, then

$\overrightarrow{\mathrm{AB}}=k\overrightarrow{\mathrm{AC}}$ for some scalar k

$$\begin{gathered} \Rightarrow \left(1-\lambda \right)\hat{i}+9\hat{j}=k\left[\left(3-\lambda \right)\hat{i}+15\hat{j}\right] \\ \Rightarrow 1-\lambda =k\left(3-\lambda \right)\mathrm{and}9=15k\left(\mathrm{Equating}\mathrm{coefficients}\mathrm{of}\hat{i}\mathrm{and}\hat{j}\right) \\ \Rightarrow 1-\lambda =\frac{3}{5}\left(3-\lambda \right) \\ \Rightarrow 5-5\lambda =9-3\lambda \\ \Rightarrow 2\lambda =-4 \\ \Rightarrow \lambda =-2 \end{gathered}$$

Thus, the value of λ is −2.