Question

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

Answer 1

Here, we will substitute the given vectors in the formula of the equation of a line that passes through a point and is parallel to a line to find the required equation. Now, in order to find the position vector of $P$, we will use the distance formula and the direction ratios from the equation found, which will help us to find the required position vector of point $P$.

Formula Used:

1. The vector equation of a line which passes through a given point and is parallel to a given line is: $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$
2. Distance formula, $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} = d$

Complete step by step solution:
The given line passes through the point $A$ whose position vector is $3i + j - k$
Let the vector $\overrightarrow a = 3i + j - k$
Also, the given line is parallel to the vector $2i - j + 2k$
Hence, let this vector be $\overrightarrow b = 2i - j + 2k$
Now, the vector equation of a line which passes through a given point and is parallel to a given line is:
$\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$
Here, substituting the above values we get,
$\Rightarrow \overrightarrow r = \left( {3i + j - k} \right) + \lambda \left( {2i - j + 2k} \right)$
Hence, the equation of the line parallel to the vector $2i - j + 2k$ and which passes through the point $A$ whose position vector is $3i + j - k$ is:
$\Rightarrow \overrightarrow r = \left( {3i + j - k} \right) + \lambda \left( {2i - j + 2k} \right)$
Now, let the coordinates of point $P = \left( {a,b,c} \right)$
And the given coordinates of point $A = \left( {3,1, - 1} \right)$
Therefore, the direction ratios of $AP = \left( {a - 3,b - 1,c + 1} \right)$
But it is given that $AP$ is parallel to $2i - j + 2k$
Hence, we can write it as:
$\dfrac{{a - 3}}{2} = \dfrac{{b - 1}}{{ - 1}} = \dfrac{{c + 1}}{2} = k$
Therefore, simplifying the above equation, we get
$a = 2k + 3$
$b = 1 - k$
And $c = 2k - 1$
Also, $AP = 15$
Hence, using distance formula $\sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2} + {{\left( {{z_2} - {z_1}} \right)}^2}} = d$, we get
$\sqrt {{{\left( {a - 3} \right)}^2} + {{\left( {b - 1} \right)}^2} + {{\left( {c + 1} \right)}^2}} = 15$
Now, substituting the values of $a,b,c$ from above, we get,
$\sqrt {{{\left( {2k + 3 - 3} \right)}^2} + {{\left( {k - 1 - 1} \right)}^2} + {{\left( {2k - 1 + 1} \right)}^2}} = 15$
$\Rightarrow \sqrt {4{k^2} + {k^2} + 4{k^2}} = 15$
Solving further, we get
$\Rightarrow \sqrt {9{k^2}} = 15$
$\Rightarrow 3k = 15$
Dividing both sides by 3, we get
$\Rightarrow k = 5$
Hence, substituting this value, we get,
$a = 2 \times 5 + 3 = 13$
$b = 1 - 5 = - 4$
And $c = 2\left( 5 \right) - 1 = 9$

**Therefore, the position vector of point $P = 13i - 4j + 9k$
**

Note:
A scalar is a quantity that has a magnitude whereas; a vector is a mathematical quantity that has both magnitude and direction. A line of given length and pointing along a given direction, such as an arrow, is a typical representation of a vector. Now, position vector is also known as location vector, it is a straight line having one end fixed and the other end attached to a moving point, it is used to describe the position of a certain point, which turns out to be its respective coordinates.