If the distance of the point P(43, , ),

Question

If the distance of the point $P(43,\alpha ,\beta )$, $\beta <0$, from the line $\vec{r}=4\hat{i}-\hat{k}+\mu (2\hat{i}+3\hat{k}),\mu \in R$ along a line with direction ratios $3,-1,0$ is $13\sqrt{10}$, then ${\alpha }^{2}+{\beta }^{2}$ is equal to \_\_\_\_\_\_\_\_

Answer: 170

Accepted Answer

Point $P$ : $(43,\alpha ,\beta )$ with the constraint $\beta

Target Line ( ${L}\_{1}$ ) : Given by $\vec{r}=4\overset{\^}{i}-\overset{\^}{k}+\mu (2\overset{\^}{i}+3\overset{\^}{k})$.

A general point $Q$ on this line is $(4+2\mu ,0,-1+3\mu )$.

Direction of measurement : The distance is measured along a line with direction ratios $(3,-1,0)$.

Find Point $Q$ on the Target Line

Since the distance is measured along the direction $d=(3,-1,0)$, the vector $PQ$ must be parallel to $d$.

$$$\begin{gathered} PQ=Q-P=(4+2\mu -43,\, 0-\alpha ,\, -1+3\mu -\beta ) \ PQ=(2\mu -39,\, -\alpha ,\, 3\mu -\beta -1) \ PQ=(2\mu -39,\, -\alpha ,\, 3\mu -\beta -1) \end{gathered}$$$

Because $PQ$ is parallel to $(3,-1,0)$, their components must be proportional :

1.$\frac{2\mu -39}{3}=\frac{-\alpha }{-1}\implies 2\mu -39=3\alpha \implies 2\mu =3\alpha +39$

2.$\frac{3\mu -\beta -1}{0}$ suggests the numerator must be zero for a finite ratio to exist: $3\mu -\beta -1=0\implies \beta =3\mu -1$

Calculate the Distance $PQ$

The distance $PQ$ is given as $13\sqrt{10}$. Using the distance formula on the proportional components where the ratio is $k$ :

$$$\begin{gathered} PQ=k(3,-1,0)=(3k,-k,0) \ |PQ|=\sqrt{(3k{)}^{2}+(-k{)}^{2}+{0}^{2}}=\sqrt{10{k}^{2}}=|k|\sqrt{10} \end{gathered}$$$

Given $|PQ|=13\sqrt{10}$, we find $|k|=13$, so $k=\pm 13$.

Solve for $\alpha ,\mu $, and $\beta $

From our proportionality equations :

$-\alpha =-k\implies \alpha =k$

$2\mu -39=3k$

Case 1 : $k=13$

$\alpha =13$

$2\mu -39=3(13)=39\implies 2\mu =78\implies \mu =39$

$\beta =3\mu -1=3(39)-1=116$

Constraint Check : $\beta

Case 2 : $k=-13$

$\alpha =-13$

$2\mu -39=3(-13)=-39\implies 2\mu =0\implies \mu =0$

$\beta =3(0)-1=-1$

Constraint Check : $\beta

${\alpha }\^{2}+{\beta }\^{2}=(-13{)}\^{2}+(-1{)}\^{2}=169+1=170$

The value of $

Solutions · 1