Question

If the points $(\alpha ,-1),(2,1)$ and $(4,5)$ are collinear, then find $\alpha$ by vector method.

• A. $4$
• B. $1$
• C. $8$
• D. None of these

Answer 1

The correct option is B $1$

If there points are collinear then vectors from one to another will have scalar triple produced $0$.Point $(\alpha ,-1),(2,1),(4,5)$

$(2-\alpha )\hat{i}+2\hat{j}-\overline{A}$

$2\hat{i}+4\hat{j}-\overline{B}$

$(4-\alpha )\hat{i}+6\hat{j}-\overline{C}$

$\overline{A}.(\overline{B}\times \overline{C})=0$

$((2-\alpha )\hat{i}+2\hat{j})(2\hat{i}+4\hat{j})\times ((4-\alpha )\hat{i}+6\hat{j})((2-\alpha )\hat{i}+2\hat{j}).[12\hat{k}-16\hat{k}+4\alpha \hat{k}]=0$

$4\alpha =4$

$\alpha =1$

Also the direction vector will be proportion

$(2-\alpha ,2)=\lambda (4-2.5-1)$

$(2-\alpha ,2)=\lambda (2,4)$

$\lambda =\frac{1}{2}$ as $2=4\lambda$

$2-\alpha =1$

$\therefore \alpha =1$