Question

Find the value of ‘p’ for which the vectors \[\vec a = 3\hat i + 2\hat j + 9\hat k\] and \[\vec b = \hat i + p\hat j + 3\hat k\] are parallel.

Answer 1

Hint: To solve this problem use the property of the parallel vector. The parallel vectors have the same direction. The components of one vector must be in the same ratio to the corresponding components of the parallel vector.
If the two vectors \[\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\] and \[\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\] are parallel then,
$\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}} = \dfrac{{{a_3}}}{{{b_3}}}$
Substitute the value , \[{a_1} = 3,{a_2} = 2,{a_3} = 9\]and \[{b_1} = 1,{b_2} = p,{b_3} = 3\]into $\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}} = \dfrac{{{a_3}}}{{{b_3}}}$ . Solve the two ratios which gives the value of $p$.

Complete step-by-step solution:
Consider the two vectors \[\vec a = 3\hat i + 2\hat j + 9\hat k\] and \[\vec b = \hat i + p\hat j + 3\hat k\]. We know that, If the two vectors \[\vec a = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\] and \[\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\] are parallel then,
$\dfrac{{{a_1}}}{{{b_1}}} = \dfrac{{{a_2}}}{{{b_2}}} = \dfrac{{{a_3}}}{{{b_3}}}$
Substitute the value , \[{a_1} = 3,{a_2} = 2,{a_3} = 9\]and \[{b_1} = 1,{b_2} = p,{b_3} = 3\]into the above relation.
$ \Rightarrow \dfrac{3}{1} = \dfrac{2}{p} = \dfrac{9}{3}$
Solve the ratio, $\dfrac{3}{1} = \dfrac{2}{p}$ we get,
$p = \dfrac{2}{3}$

The value of ‘p’ for which the vectors \[\vec a = 3\hat i + 2\hat j + 9\hat k\] and \[\vec b = \hat i + p\hat j + 3\hat k\] are parallel is $\dfrac{2}{3}$.

Note: Two vectors are parallel if they are scalar multiples of one another.
If u and v are two non-zero vectors and $u = \lambda v$ , then u and v are parallel.
The two vectors \[\vec a = 3\hat i + 2\hat j + 9\hat k\] and \[\vec b = \hat i + p\hat j + 3\hat k\] are parallel then,
\[3\hat i + 2\hat j + 9\hat k = \lambda (\hat i + p\hat j + 3\hat k)\]
$ \Rightarrow 3\hat i + 2\hat j + 9\hat k = \lambda \hat i + p\lambda \hat j + 3\lambda \hat k$
\[ \Rightarrow 3 = \lambda ,2 = p\lambda ,9 = 3\lambda \]
Substitute $\lambda = 3$ into \[2 = p\lambda \] we get,
\[2 = 3p\]
$ \Rightarrow p = \dfrac{2}{3}$