Q.Find the position vector of the mid point of the vector joining the points $P(2, 3, 4)$ and $Q(4, 1, -2)$.

Concept and Intuition

The internal section formula: a point dividing $PQ$ in ratio $m:n$ has position vector $\dfrac{n\,\vec P+m\,\vec Q}{m+n}$.

Step-by-Step Solution

1. Here $m:n=1:2$ from P to Q, so $\vec R=\dfrac{2\,\overrightarrow{OP}+1\,\overrightarrow{OQ}}{3}$.
2. Substitute: $2(2\vec a-\vec b)+(\vec a+3\vec b)=4\vec a-2\vec b+\vec a+3\vec b=5\vec a+\vec b$.
3. Divide by 3: $\vec R=\dfrac13(5\vec a+\vec b)$. …

$A=(2,-\lambda,5)$, $B=(\mu,7,3)$. Midpoint $=\left(\dfrac{2+\mu}{2},\dfrac{-\lambda+7}{2},\dfrac{8}{2}\right)=(3,2,4)$. …

Concept and Intuition

The condition $2|\overrightarrow{PR}|=|\overrightarrow{RQ}|$ means $|PR|:|RQ|=1:2$, so R divides segment PQ internally in the ratio $1:2$ measured from P.

Step-by-Step Solution

1. Section formula: $R=P+\tfrac{1}{1+2}(Q-P)=P+\tfrac13(Q-P)$.
2. $Q-P=(8-2,-3-6,1-4)=(6,-9,-3)$. …

Position vectors: $\vec A=2\vec a+6\vec b$, $\vec B=3\vec a-7\vec b$. $P$ divides $AB$ internally in ratio $3:5$, so $AP:PB=3:5$ and

Then

$\vec{PB}=\vec B-\vec P=\vec B-\vec A-\frac38(\vec B-\vec A)=\frac58(\vec B-\vec A).$ …

Let $R$ divide $PQ$ internally in ratio $m:n$. Then

$R=\frac{m\,\vec{Q}+n\,\vec{P}}{m+n}=\frac{m(\vec a+2\vec b)+n(5\vec a-6\vec b)}{m+n}=\frac{(m+5n)\vec a+(2m-6n)\vec b}{m+n}.$ …

Midpoint of $A(-4,3)$ and $B(0,5)$ is $M=(-2,4)$.

Slope of line through $C(-1,2)$ and $M(-2,4)$:

$m=\frac{4-2}{-2-(-1)}=\frac{2}{-1}=-2.$ …

The position vector of $P$ is $\vec{OP}=\dfrac{m\vec b+n\vec a}{m+n}$. Then …

Reflecting $(1,2)$ across the $x$-axis gives $(1,-2)$. The reflected path is the straight line from $(1,-2)$ to $(5,6)$:

$\text{slope}=\frac{6-(-2)}{5-1}=\frac{8}{4}=2.$ …

Concept and Intuition

The xz-plane is $y=0$. Use the section formula on the $y$-coordinate only.

Step-by-Step Solution

1. Dividing $(2,4,7)$ and $(3,-5,8)$ in ratio $\alpha:1$, the $y$-coordinate is $\dfrac{\alpha(-5)+1(4)}{\alpha+1}$.
2. Set to $0$: $-5\alpha+4=0$.
3. $\alpha=\tfrac45$. …

Concept and Intuition

The segment $PQ$ has slope $-a/b$ (parallel to $ax+by=0$). Use its midpoint $(1,5)$ with $P,Q$ on the two given lines to find the slope.

Step-by-Step Solution

1. Let $P=(p,5p+4)$ on $5x-y+4=0$ and $Q=(q,\tfrac{4-3q}{4})$ on $3x+4y-4=0$.
2. Midpoint: $p+q=2$ and $20p-3q=20$, giving $p=\tfrac{26}{23},q=\tfrac{20}{23}$.
3. Slope of $PQ=\dfrac{222/23-8/23}{26/23-20/23}=\dfrac{214}{6}=\dfrac{107}{3}$. …

Let $R=(a,b)$ be a point on the parabola, so $b^2=2a$.

The midpoint $M=(h,k)$ of segment $OR$ (with $O=(0,0)$) is

$h=\frac{a}{2},\qquad k=\frac{b}{2}\quad\Rightarrow\quad a=2h,\ b=2k.$ …

Concept and Intuition

All three points are collinear with $B$ between $A$ and $C$; the ratio of lengths translates into a scalar multiple of the direction vector $\vec{AB}$.

Step-by-Step Solution

1. $\vec{AB}=B-A=(1-(-1),\,3-2)=(2,1)$.
2. $BC=8\,AB$ and $B$ on segment $AC$ so $AC=AB+BC=9\,AB$.
3. $\vec{AC}=9(2,1)=(18,9)$.
4. $C=A+\vec{AC}=(-1+18,\,2+9)=(17,11)$. …