Question

The two vectors \[a\] and \[b\] are perpendicular. If \[a\] has magnitude 8 and \[b\] has magnitude 3. What is \[\left| {a - 2b} \right|\]?

Answer 1

Hint: First we will take square of the given expression \[\left| {a - 2b} \right|\]. Then we will use the rule \[\vec a \cdot \vec a = {\left| {\vec a} \right|^2}\] and \[\vec b \cdot \vec a = \vec a \cdot \vec b\] in the obtained equation. Then simplify it to find the required value.

Complete step-by-step answer:
We are given that the two vectors \[a\] and \[b\] are perpendicular, \[\left| {\vec a} \right| = 8\] and \[\left| {\vec b} \right| = 3\].
Since we know that \[\vec a\] and \[\vec b\] are perpendicular, so \[\vec a \cdot \vec b = 0\].
Taking square of the given expression \[\left| {a - 2b} \right|\], we get
\[ \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = \left( {\vec a - 2\vec b} \right) \cdot \left( {\vec a - 2\vec b} \right)\]
Simplifying the right hand side of the above equation, we get
\[ \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = \vec a \cdot \vec a - 2 \cdot \vec b \cdot \vec a - 2 \cdot \vec a \cdot \vec b + 4\vec b \cdot \vec b\]
Using the rule, \[\vec a \cdot \vec a = {\left| {\vec a} \right|^2}\] in the above equation, we get
\[ \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = {\left| {\vec a} \right|^2} - 2 \cdot \vec b \cdot \vec a - 2 \cdot \vec a \cdot \vec b + 4{\left| {\vec b} \right|^2}\]
Using the rule, \[\vec b \cdot \vec a = \vec a \cdot \vec b\] in the above equation, we get
\[
   \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = {\left| {\vec a} \right|^2} - 2 \cdot \vec a \cdot \vec b - 2 \cdot \vec a \cdot \vec b + 4{\left| {\vec b} \right|^2} \\
   \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = {\left| {\vec a} \right|^2} - 4\vec a \cdot \vec b + 4{\left| {\vec b} \right|^2} \\
 \]
Substituting the values \[\left| {\vec a} \right| = 8\] , \[\left| {\vec b} \right| = 3\] and \[\vec a \cdot \vec b = 0\] in the above equation, we get
\[
   \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = {8^2} - 4\left( 0 \right) + 4\left( {{3^2}} \right) \\
   \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = 64 - 0 + 4 \cdot 9 \\
   \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = 64 + 36 \\
   \Rightarrow {\left| {\vec a - 2\vec b} \right|^2} = 100 \\
 \]
Taking square root on both sides of the above equation, we get
\[
   \Rightarrow \left| {\vec a - 2\vec b} \right| = \pm \sqrt {100} \\
   \Rightarrow \left| {\vec a - 2\vec b} \right| = \pm 10 \\
 \]
Since the magnitude can never be negative, the negative value of \[\left| {\vec a - 2\vec b} \right|\] is discarded.
Therefore, 10 is the required value.

Note: One should know that the magnitude of a vector is the length of a line segment and the vector, which has a magnitude of 1 is known as the unit vector. The key point is to use the rules \[\vec a \cdot \vec a = {\left| {\vec a} \right|^2}\] and \[\vec b \cdot \vec a = \vec a \cdot \vec b\] to simplify. We need to know that when two vectors are perpendicular, then their dot product is always zero or else the answer will be wrong.