Question

Find the vector equation of a line passing through the point A whose position vector is $ 3\overline i + \overline j - \overline k $ and which is parallel to the vector $ 2\overline i - \overline j + 2\overline {k.} $ If P is a point of this line such that $ AP = 15, $ find the position vector of P.

Answer 1

Hint: First find the parallel point assuming a variable “r” with respect to the point and the vector position parallel to the other vector and then find the value of “r” using the distance formula. Replace the value of “r” in the assumed vector point.

Complete step-by-step answer:
Equation of the line passing through the point A whose position vector is $ 3\overline i + \overline j - \overline k $ and which is parallel to the vector $ 2\overline i - \overline j + 2\overline {k.} $ is
 $ $ $ \dfrac{{x - 3}}{2} = \dfrac{{y - 1}}{{ - 1}} = \dfrac{{z + 1}}{2} $
Let the point A be $ (3,1, - 1) $
And the parallel point P be $ (2r + 3, - r + 1,2r - 1) $ .... (A)
Also, given that $ AP = 15 $
Now, using the distance formula –
AP $ = \sqrt {{{(2r + 3 - 3)}^2} + {{( - r + 1 - 1)}^2} + {{(2r - 1 + 1)}^2}} $
Simplify the equation –
AP $ = \sqrt {{{(2r)}^2} + {{( - r)}^2} + {{(2r)}^2}} $
Place the given value on the left hand side of the equation –
 $ 15 = \sqrt {{{(2r)}^2} + {{( - r)}^2} + {{(2r)}^2}} $
Take square on both the sides of the above equation –
 \[{15^2} = {\left( {\sqrt {{{(2r)}^2} + {{( - r)}^2} + {{(2r)}^2}} } \right)^2}\]
Square and square-root cancel each other on the right hand side of the equation –
 \[225 = {(2r)^2} + {( - r)^2} + {(2r)^2}\]
Simplify the above equation. Square of the negative term also gives the positive term.
 \[ \Rightarrow 225 = 4{r^2} + {r^2} + 4{r^2}\]
Simplify –
 \[ \Rightarrow 225 = 9{r^2}\]
When the term multiplicative on one side is moved to the opposite side then it goes to the denominator on the opposite side.
 \[ \Rightarrow \dfrac{{225}}{9} = {r^2}\]
Find factors on the numerator part of the fraction-
 \[ \Rightarrow \dfrac{{25 \times 9}}{9} = {r^2}\]
Common multiple from the numerator and the denominator cancel each other. Therefore remove from the numerator and the denominator.
 \[ \Rightarrow 25 = {r^2}\]
The above equation can be re-written as –
 \[ \Rightarrow {r^2} = 25\]
Find the square-root on both the sides of the equation –
 \[ \Rightarrow r = \pm 5\]
Now, place values in equation (A)
P $ (2r + 3, - r + 1,2r - 1) $
When $ r = 5 $
P $ (2(5) + 3, - 5 + 1,2(5) - 1) = P(13, - 4,9) $ ... (B)
When $ r = - 5 $
P $ (2( - 5) + 3, - ( - 5) + 1,2( - 5) - 1) = P( - 7,6, - 11) $ ....(C)
Hence, the position vectors of P are $ (13, - 4,9){\text{ and ( - 7,6, - 11)}} $
So, the correct answer is “ $ (13, - 4,9){\text{ and ( - 7,6, - 11)}} $ ”.

Note: Be careful while simplifying the points. Remember when there is product of minus with minus it gives plus positive resultant value and when there is product of one negative and positive value then the resultant value is negative. Also remember the distance formula.