Question

If \({\bf{a}} = \langle 3,0, - 1\rangle \), find a vector \({\bf{b}}\) such that \({{\mathop{\rm comp}\nolimits} _{\bf{a}}}{\bf{b}} = 2\).

Short answer

Vector \({\bf{b}}\) is \(\langle s,t,3s - 2\sqrt {10} \rangle \)

Step-by-step solution

Formula and concept used for finding vector \({\bf{b}}\)

The expression for scalar projection of\({\bf{b}}\)onto\(a\)is:

\({{\mathop{\rm comp}\nolimits} _{\rm{a}}}{\rm{b}} = \frac{{{\rm{a}} \cdot {\rm{b}}}}{{|{\rm{a}}|}}\). …… (1)

Consider a general expression to find the dot product between two three-dimensional vectors.

\({\rm{a}} \cdot {\rm{b}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \cdot \left\langle {{b_1},{b_2},{b_3}} \right\rangle \)

\({\rm{a}} \cdot {\rm{b}} = {a_1}{b_1} + {a_2}{b_2} + {a_3}{b_3}\) …… (2)

Consider a general expression to find the magnitude of a three dimensional vector that is\({\rm{a}} = \left\langle {{a_1},{a_2},{a_3}} \right\rangle \).

\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + + a_3^2} \) …… (3)

Calculation for finding the vector \({\bf{b}}\)

Given:

\({\rm{a}} = \langle 3,0, - 1\rangle \)and \({{\mathop{\rm comp}\nolimits} _{\rm{a}}}{\rm{b}} = 2.\)

In equation (3), substitute 3 for \({a_1},0\) for \({a_2}\) and \( - 1\) for \({a_3}\).

\(\begin{aligned}{l}|a| &= \sqrt {{{(3)}^2} + {{(0)}^2} + {{( - 1)}^2}} \\ &= \sqrt {9 + 0 + 1} \\ &= \sqrt {10} \end{aligned}\)

In equation (1), substitute \(2\) for \({{\mathop{\rm comp}\nolimits} _{\rm{a}}}{\rm{b}}\) and \(\sqrt {10} \) for \(|{\rm{a}}|\).

\(\begin{aligned}{l}2 &= \frac{{a \cdot b}}{{\sqrt {10} }}\\a \cdot b &= 2\sqrt {10} \end{aligned}\)

In equation (2), substitute \(3\) for \({a_1}\), \(0\) for \({a_2}\), \( - 1\) for \({a_3}\) and \(2\sqrt {10} \) for \({\rm{a}} \cdot {\rm{b}}\).

\(\begin{aligned}{l}2\sqrt {10} &= (3){b_1} + (0){b_2} + ( - 1){b_3}\\2\sqrt {10} &= 3{b_1} - {b_3}\end{aligned}\) …… (4)

In general, there are equations required to obtain the values of three variables \(\left( {{b_1},{b_2},{b_3}} \right)\). But in this case, there is only one equation available. So, multiple solutions are possible.

Consider \({b_1} = s\) and \({b_2} = t\)

In equation (4), substitute \(s\) for \({b_1}\).

\(\begin{aligned}{l}2\sqrt {10} &= 3(s) - {b_3}\\{b_3} &= 3s - 2\sqrt {10} \end{aligned}\)

Thus, the vector \({\rm{b}}\) is \(\langle s,t,3s - 2\sqrt {10} \rangle \).