Question

To determine the values of \(a + b,2a + 3b,|a|\), and \(|a - b|\)

Short answer

The value of the vector \(a + b\) is \(2{\rm{i}} - 2{\rm{j}} + 3{\bf{k}}\)

The value of the vector \(2{\rm{a}} + 3\;{\rm{b}}\) is \(4{\rm{i}} - 2{\rm{j}} + 5{\rm{k}}\)

The value of the vector \(|a|\)is \(6\) .

The value of the vector \(|a - b|\) is \(\sqrt {65} \).

Step-by-step solution

Concept of Arithmetical Operation on Vectors

Consider the two two-dimensional vectors such as\(a = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}\)and\({b_1}{\bf{i}} + {b_2}{\bf{j}} + {b_3}{\bf{k}}\).

1. The vector sum of two vectors\(a\)and\(b\)is\(a + b = \left( {{a_1} + {b_1}} \right){\bf{i}} + \left( {{a_2} + {b_2}} \right){\bf{j}} + \left( {{a_3} + {b_3}} \right){\bf{k}}\).

2. The vector subtraction of two vectors\(a\)and\(b\)is\((a - b) = \left( {{a_1} - {b_1}} \right){\bf{i}} + \left( {{a_2} - {b_2}} \right){\bf{j}} + \left( {{a_3} - {b_3}} \right){\bf{k}}\).

3. The constant multiplication of vector\(a\)is\(ca = c{a_1}{\bf{i}} + c{a_2}{\bf{j}} + c{a_3}{\bf{k}}\).

Calculation of the vector\(a + b\)

The given vectors are \(a = 2{\bf{i}} - 4{\bf{j}} + 4{\bf{k}}\) and \({\bf{b}} = 2{\rm{j}} - {\bf{k}}\).

From the above vectors the values of components are \({a_1} = 2,{a_2} = - 4,{a_3} = 4,{b_1} = 0,{b_2} = 2\), and \({b_3} = - 1\).

Use the formula with \({a_1} = 2,{a_2} = - 4,{a_3} = 4,{b_1} = 0,{b_2} = 2\), and \({b_3} = - 1\) to obtain \(a + b.\)

\(\begin{aligned}{l}a + b = 2{\bf{i}} - 4{\bf{j}} + 4{\bf{k}} + (2{\bf{j}} - {\bf{k}})\\a + b = (2 + 0){\rm{i}} + ( - 4 + 2){\bf{j}} + (4 - 1){\bf{k}}\\a + b = 2{\rm{i}} - 2{\rm{j}} + 3{\bf{k}}\end{aligned}\)

Thus, the value of the vector v is \(2i - 2j + 3k\).

Calculation of the vector\(2{\rm{a}} + 3\;{\rm{b}}\)

From formula, write the expression to find \(2{\rm{a}} + 3\;{\rm{b}}\).

Use the formula with \({a_1} = 2,{a_2} = - 4,{a_3} = 4,{b_1} = 0,{b_2} = 2\), and \({b_3} = - 1\) to obtain \(2{\rm{a}} + 3\;{\rm{b}}.\)\(\begin{aligned}{l}2{\rm{a}} + 3\;{\rm{b}} = (2(2) + 3(0)){\rm{i}} + (2( - 4) + 3(2)){\rm{j}} + (2(4) + 3( - 1)){\rm{j}}\\2{\rm{a}} + 3\;{\rm{b}} = (4 + 0){\rm{i}} + ( - 8 + 6){\rm{j}} + (8 - 3){\rm{k}}\\2{\rm{a}} + 3\;{\rm{b}} = 4{\rm{i}} - 2{\rm{j}} + 5{\rm{k}}\end{aligned}\)

Thus, the value of the vector \(2{\rm{a}} + 3\;{\rm{b}}\) is \(4{\rm{i}} - 2{\rm{j}} + 5{\rm{k}}\).

Calculation of the vector\(|a|\)

Formula used:

The formula for the magnitude of vector\(a = {a_1}{\bf{i}} + {a_2}{\bf{j}} + {a_3}{\bf{k}}\)is\(|{\rm{a}}| = \sqrt {a_1^2 + a_2^2 + + a_3^2} \).

Use the formula with \({a_1} = 2,{a_2} = - 4\), and \({a_3} = 4\) to obtain the magnitude of \(|{\rm{a}}|\).

\(\begin{aligned}{l}|a| = \sqrt {{2^2} + {{( - 4)}^2} + {4^2}} \\|a| = \sqrt {4 + 16 + 16} \\|a| = \sqrt {36} \\|a| = 6\end{aligned}\)

Thus, the value of the vector \(|{\rm{a}}|\) is\(6\) .

Calculation of the vector\(|a - b|\)

Use the formula with \({a_1} = 2,{a_2} = - 4,{a_3} = 4,{b_1} = 0,{b_2} = 2\), and \({b_3} = - 1\) to find difference \(a - b\).

\(\begin{aligned}{l}a - b = 2{\bf{i}} - 4{\bf{j}} + 4{\bf{k}} - (2{\bf{j}} - {\bf{k}})\\a - b = (2 - 0){\rm{i}} + ( - 4 - 2){\bf{j}} + (4 + 1){\bf{k}}\\a - b = 2{\rm{i}} - 6{\bf{j}} + 5{\bf{k}}\end{aligned}\)

Hence, the difference of the two vector value is \(a - b = 2{\bf{i}} - 6{\bf{j}} + 5{\bf{k}}\).

Use the formula with \({a_1} = 2,{a_2} = - 6\), and \({a_3} = 5\) to obtain the magnitude of v.

\(\begin{aligned}{l}|a - b| = \sqrt {{2^2} + {{( - 6)}^2} + {5^2}} \\|a - b| = \sqrt {4 + 36 + 25} \\|a - b| = \sqrt {65} \end{aligned}\)

Thus, the value of the vector \(|a - b|\) is \(\sqrt {65} \).