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

Last updated date: 24th Jun 2024
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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.

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.