Question

If the line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$, meet the plane x + 2y + 3z = 15 at the point P, then the distance of P from the origin is
(a) $\dfrac{9}{2}$
(a) $2\sqrt{5}$
(a) $\dfrac{\sqrt{5}}{2}$
(a) $\dfrac{7}{2}$

Answer 1

Hint: To solve this question, first we will find the values of x, y and x in terms of k, which we will get from equation $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=k$. Then, we know that point P lies on both $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=k$ and plane x + 2y + 3z = 15, then we will put values of x, y and z in plane x + 2y + 3z = 15 and we will get value of k and then we will evaluate point P and also distance of Point P from origin.

Complete step by step answer:
We know that, the standard form of equation $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line.
So, for $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$, we have 2, 3, 4 as direction ratios and ( 1, -1, 2 ) point on line.
Let say $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}=k$
Then any point on line $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ will be of form P( 2k + 1, 3k – 1, 4k + 2 ) where, we have point x =2k + 1, y = 3k – 1, z = 4k + 2.
Now, in question it is given that point P lies on plane x + 2y + 3z = 15
Putting values of x =2k + 1, y = 3k – 1, z = 4k + 2 in x + 2y + 3z = 15, as point P will satisfy the equation of plane, we get
(2k + 1)+ 2(3k – 1) + 3(4k + 2) = 15
On solving we get
2k + 6k + 12k + 1 -2 + 6 = 15
20k + 5 = 15
20k = 10
On solving, we get
$k=\dfrac{1}{2}$
Putting values of $k=\dfrac{1}{2}$ in P( 2k + 1, 3k – 1, 4k + 2 ), we get
$P\left( 2\cdot \dfrac{1}{2}+1,3\cdot \dfrac{1}{2}-1,4\cdot \dfrac{1}{2}+2 \right)$
On solving we get,
\[P\left( 2,\dfrac{1}{2},4 \right)\]
Now, we know that distance of any point P ( x, y, z ) from origin is equals to $\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$
So, distance of \[P\left( 2,\dfrac{1}{2},4 \right)\] from ( 0,0,0 ) is $\sqrt{{{2}^{2}}+{{\left( \dfrac{1}{2} \right)}^{2}}+{{4}^{2}}}$
On solving, we get
$=\sqrt{4+\left( \dfrac{1}{4} \right)+16}$
$=\sqrt{\left( \dfrac{81}{4} \right)}$
$=\dfrac{9}{2}$units

So, the correct answer is “Option A”.

Note: Always convert the equation of line firstly to standard line equation which is $\dfrac{x-a}{d}=\dfrac{y-b}{e}=\dfrac{z-c}{f}$, where d, e, f are direction ratios and ( a, b, c ) lying on line. Always remember that distance of any point P ( x, y, z ) from origin is equals to $\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$. Avoid calculation error while solving the question.