Question

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

Answer 1

Let the points are A(k, -10, 3) , B (1, -1, 3) and C (3, 5, 3 )

$\text{So,}\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(\hat{i}-\hat{j}+3\hat{k})-(k\hat{i}-10\hat{j}+3\hat{k})=(1-k)\hat{i}+(-1+10)\hat{j}+(3-3)\hat{k}=(1-k)\hat{k}+9\hat{j}+0\hat{k}\therefore |\overrightarrow{AB}|=\sqrt{(1-k{)}^2+(9{)}^2+0}=\sqrt{(1-k{)}^2+81}\text{Similarly,}\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(3\hat{i}+5\hat{j}+3\hat{k})-(k\hat{k}-10\hat{j}+3\hat{k}=2\widehat{(}i)+6\hat{j}+0\hat{k}\therefore |\overrightarrow{BC}|=\sqrt{2^2+6^2+0}=2\sqrt{10}\text{and}\overrightarrow{AC}=\overrightarrow{OC}-\overrightarrow{OA}=(3\hat{i}+5\hat{j}+3\hat{k})-(k\hat{i}-10\hat{j}+3\hat{k})=(3-k)\hat{i}+15\hat{j}+0\hat{k}\therefore |\overrightarrow{AC}|=\sqrt{(3-k{)}^2+225}$

If A, B and C are collinear, then sum of modulus of any two vectors will be equal to the modulus of third vectors

$\text{For}|\overrightarrow{AB}|+|\overrightarrow{BC}|=|\overrightarrow{AC}|,\sqrt{(1-k{)}^2+81}+2\sqrt{10}=\sqrt{(3-k{)}^2+225}\Rightarrow \sqrt{(3-k{)}^2+225}-\sqrt{(1-k{)}^2+81}=2\sqrt{10}\Rightarrow \sqrt{9+k^2-6k+225}-\sqrt{1+k^2-2k+81}=2\sqrt{10}\Rightarrow \sqrt{k^2-6k+234}-2\sqrt{10}=\sqrt{k^2-2k+82}\Rightarrow k^2-6k+234+40-2\sqrt{k^2-6k+234}.2\sqrt{10}=k^2-2k+82\Rightarrow k^2-6k+234+40-k^2+2k-82=4\sqrt{10}\sqrt{k^2+234-6k}\Rightarrow -4k+192=4\sqrt{10}\sqrt{k^2+234-6k}\Rightarrow -k+48=\sqrt{10}\sqrt{k^2+234-6k}$

On squaring both sides, we get

$48\times 48+k^2-96k=10(k^2+234-6k)$

$\Rightarrow k^2-96k-10k^2+60k=-48\times 48+2340$

$\Rightarrow -9k^2-36k=-48\times 48+2340$

$\Rightarrow (k^2+4k)=+16\times 16-260$ [dividing by 9 in both sides]

$\Rightarrow k^2+4k=-4$

$k^2+4k+4=0$

$\Rightarrow (k+2{)}^2=0$

$\therefore k=-2$