Question

$A=(1,2,3),B=(4,5,7),C=(-4,3,-6),D=(2,k,2)$ are four points. If the lines $AB$ and $CD$ are parallel, then $k=$

• A. $0$
• B. $-9$
• C. $9$
• D. $2$

Answer 1

The correct option is C $9$

Given, $A=(1,2,3),B=(4,5,7),C=(-4,3,-6),D=(2,k,2)$
For $AB$ to be parallel to $CD$, their direction cosines should be equal.
Therefore, Dcs of $AB$ will be
$(\frac{3}{\sqrt{34}},\frac{3}{\sqrt{34}},\frac{4}{\sqrt{34}})$

D.C.s of $CD$ will be
$(\frac{6}{\sqrt{100+(k-3{)}^2}},\frac{k-3}{\sqrt{100+(k-3{)}^2}},\frac{8}{\sqrt{100+(k-3{)}^2}})$
Now $\frac{6}{\sqrt{100+(k-3{)}^2}}=\frac{3}{\sqrt{34}}$

$\frac{k-3}{\sqrt{100+(k-3{)}^2}}=\frac{3}{\sqrt{34}}$

$\frac{8}{\sqrt{100+(k-3{)}^2}}=\frac{4}{\sqrt{34}}$

Solving any of the above can give us the value of $k$:
$\frac{6}{\sqrt{100+(k-3{)}^2}}=\frac{3}{\sqrt{34}}$

$4\left(34\right)=100+(k-3{)}^2$

$136=100+(k-3{)}^2$
$36=(k-3{)}^2$
$k-3=\pm 6$
$k=9$ and $k=-3$