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

Hint:
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.