Question

If \((a,b,c)\) are the direction ratios of a line joining the points \((4,3,-5)\) and \((-2,1,-8)\), then the point \(P(a,3b,2c)\) lies on the plane

• A. \(x+y+z=0\)
• B. \(x+y-2z=0\)
• C. \(x+2y+3z=0\)
• D. \(x-2y+3z=0\)

Hint:

Direction ratios of the line joining \((x_1,y_1,z_1)\) and \((x_2,y_2,z_2)\) are \((x_2-x_1,\;y_2-y_1,\;z_2-z_1)\).

Answer 1

The Correct Option is B

Step 1: Recall how direction ratios of a join are found.
The direction ratios of the line joining two points are the differences of their coordinates.
Step 2: Subtract the coordinates.
For $A(4,3,-5)$ and $B(-2,1,-8)$, we get $a=-2-4=-6$, $b=1-3=-2$, $c=-8-(-5)=-3$.
Step 3: Build the point $P$.
The point is $P(a,3b,2c)=(-6,\,3(-2),\,2(-3))=(-6,-6,-6)$.
Step 4: Test the candidate planes by substitution.
A point lies on a plane when its coordinates satisfy the plane equation. We substitute $(-6,-6,-6)$.
Step 5: Check $x+y-2z=0$.
$-6+(-6)-2(-6)=-12+12=0$, which is satisfied. The other options give nonzero values, for example $x+y+z=-18\neq0$.
Step 6: Conclude.
So $P$ lies on the plane $x+y-2z=0$.
\[ \boxed{x+y-2z=0} \]