Let a triangle PQR be such that P and Q lie on the line (x+3)/(8)=(y-4)/(2)=(z+1)/(2) and …

Question

Let a triangle PQR be such that P and Q lie on the line $\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}$ and are at a distance of 6 units from $R(1,2,3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\Delta P Q R$, then $\alpha+\beta+\gamma$...

Accepted Answer

Write the given line in parametric form.

$\frac{x+3}{8}=\frac{y-4}{2}=\frac{z+1}{2}=t$

So any point on the line is

$P(t)=(-3+8t,\, 4+2t,\, -1+2t)$

Since both $P$ and $Q$ lie on this line and are at a distance $6$ from $R(1,2,3)$, we find the points on the line whose distance from $R$ is $6$.

Distance condition:

$\sqrt{(x-1{)}^{2}+(y-2{)}^{2}+(z-3{)}^{2}}=6$

Substitute $x=-3+8t,\, y=4+2t,\, z=-1+2t$:

$(-3+8t-1{)}^{2}+(4+2t-2{)}^{2}+(-1+2t-3{)}^{2}=36$

$(8t-4{)}^{2}+(2t+2{)}^{2}+(2t-4{)}^{2}=36$

Now expand:

$64{t}^{2}-64t+16+4{t}^{2}+8t+4+4{t}^{2}-16t+16=36$

$72{t}^{2}-72t+36=36$

$72{t}^{2}-72t=0$

$72t(t-1)=0$

So,

$t=0\, \, t=1$

Hence the two points are:

For $t=0$,

$P=(-3,4,-1)$

For $t=1$,

$Q=(5,6,1)$

Now centroid of triangle $PQR$ is

$(\alpha ,\beta ,\gamma )=(\frac{{x}_{P}+{x}_{Q}+{x}_{R}}{3},\frac{{y}_{P}+{y}_{Q}+{y}_{R}}{3},\frac{{z}_{P}+{z}_{Q}+{z}_{R}}{3})$

Substitute $P=(-3,4,-1)$, $Q=(5,6,1)$, $R=(1,2,3)$:

$\alpha =\frac{-3+5+1}{3}=\frac{3}{3}=1$

$\beta =\frac{4+6+2}{3}=\frac{12}{3}=4$

$\gamma =\frac{-1+1+3}{3}=\frac{3}{3}=1$

Therefore,

$\alpha +\beta +\gamma =1+4+1=6$

So the correct answer is:

$6$

Hence, Option C is correct.

Solutions · 1