Let PQR be a triangle such that PQ=-2i-j+2k and PR=ai+bj-4k,a,b Z.

Question

Let $PQR$ be a triangle such that $\overset{	o \leftarrow }{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$ and $\overset{	o \leftarrow }{PR}=a\hat{i}+b\hat{j}-4\hat{k},a,b\in Z$. Let S be the point on QR , which is equidistant from the lines PQ and PR . If $|\overset{	o \leftarrow }{PR}|=9$ and $\overset{	o \leftarrow }{PS}=\hat{i}-7\hat{j}+2\hat{k}$, then the value of $3a-4b$ is $____$

Answer: 37

Accepted Answer

We are given that $|\overset{	o \leftarrow }{PR}|=9$ Using the distance formula for the vector $PR=a\hat{i}+b\hat{j}-4\hat{k}$ :

$\sqrt{{a}^{2}+{b}^{2}+(-4{)}^{2}}=9$

$\Rightarrow$ ${a}^{2}+{b}^{2}+16=81$

$\Rightarrow$ ${a}^{2}+{b}^{2}=65$

The problem states that point $S$ lies on $QR$ and is equidistant from lines $PQ$ and $PR$. In a triangle, the locus of points equidistant from two sides is the angle bisector. Therefore, $\overset{	o \leftarrow }{PS}$ is the internal angle bisector of $\angleQPR$.

For an angle bisector $\overset{	o \leftarrow }{PS}$, it must be collinear with the sum of the unit vectors along the adjacent sides :

$\overset{	o \leftarrow }{PS}=\lambda (\frac{\overset{	o \leftarrow }{PQ}}{|\overset{	o \leftarrow }{PQ}|}+\frac{\overset{	o \leftarrow }{PR}}{|\overset{	o \leftarrow }{PR}|})$

where $\lambda $ is a scalar constant.

First, calculate $|\overset{	o \leftarrow }{PQ}|$ :

$|\overset{	o \leftarrow }{PQ}|=\sqrt{(-2{)}^{2}+(-1{)}^{2}+{2}^{2}}=\sqrt{4+1+4}=3$

Now, set up the bisector equation :

$\overset{	o \leftarrow }{PS}=\lambda (\frac{-2\hat{i}-\hat{j}+2\hat{k}}{3}+\frac{a\hat{i}+b\hat{j}-4\hat{k}}{9})$

Multiply the first term by $\frac{3}{3}$ to get a common denominator:

$\overset{	o \leftarrow }{PS}=\frac{\lambda }{9}((-6+a)\hat{i}+(-3+b)\hat{j}+(6-4)\hat{k})$

$\Rightarrow$ $\overset{	o \leftarrow }{PS}=\frac{\lambda }{9}((a-6)\hat{i}+(b-3)\hat{j}+2\hat{k})$

We are given

$\overset{	o \leftarrow }{PS}=\hat{i}-7\hat{j}+2\hat{k}$

Comparing the $\hat{k}$-components:

$\Rightarrow$ $\frac{\lambda }{9}(2)=2\Rightarrow \lambda =9$

Substitute $\lambda =9$ into other components:

For $\hat{i}$ :

$a-6=1\Rightarrow a=7$

For $\hat{j}$ :

$b-3=-7\Rightarrow b=-4$

Check against the constraint :

${a}^{2}+{b}^{2}={7}^{2}+(-4{)}^{2}=49+16=65$

The values are consistent. value of $3a-4b$ :

$3a-4b=3(7)-4(-4)$ $=21+16=37$

Solutions · 1