Question

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

Answer 1

Let the given points be A(λ, −10, 3), B(1, −1, 3) and C(3, 5, 3).
\[\overrightarrow{AB} = \left( \hat{i} - \hat{j} + 3 \hat{k} \right) - \left( \lambda \hat{i} - 10 \hat{j} + 3 \hat{k} \right) = \left( 1 - \lambda \right) \hat{i} + 9 \hat{j}\]
\[\overrightarrow{AC} = \left( 3 \hat{i} + 5 \hat{j} + 3 \hat{k} \right) - \left( \lambda \hat{i} - 10 \hat{j} + 3 \hat{k} \right) = \left( 3 - \lambda \right) \hat{i} + 15 \hat{j}\]
If the points A, B, C are collinear, then
\[\overrightarrow{AB} = k \overrightarrow{AC}\]  for some scalar k 
\[\Rightarrow \left( 1 - \lambda \right) \hat{i} + 9 \hat{j} = k\left[ \left( 3 - \lambda \right) \hat{i} + 15 \hat{j} \right]\]
\[ \Rightarrow 1 - \lambda = k\left( 3 - \lambda \right) \text{ and }9 = 15k \left(\text{ Equating coefficients of }\hat{i}\text{ and }\hat{j} \right)\]
\[ \Rightarrow 1 - \lambda = \frac{3}{5}\left( 3 - \lambda \right)\]
\[ \Rightarrow 5 - 5\lambda = 9 - 3\lambda\]
\[ \Rightarrow 2\lambda = - 4\]
\[ \Rightarrow \lambda = - 2\]
Thus, the value of λ is −2.