If $\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}$ and $\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k},$ then show that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.
We know that two nonzero vectors are perpendicular if their scalar product is zero.
First,calculate $\vec{a}+\vec{b}$:
$\vec{a}+\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k}) = 6 \hat{i}+2 \hat{j}-8 \hat{k}$
Next,calculate $\vec{a}-\vec{b}$:
$\vec{a}-\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k}) = 4 \hat{i}-4 \hat{j}+2 \hat{k}$
Now,find the dot product $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})$:
$(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = (6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot (4 \hat{i}-4 \hat{j}+2 \hat{k})$
$= (6)(4) + (2)(-4) + (-8)(2)$
$= 24 - 8 - 16$
$= 24 - 24 = 0$
Since the dot product is $0$,the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.
First,calculate $\vec{a}+\vec{b}$:
$\vec{a}+\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})+(\hat{i}+3 \hat{j}-5 \hat{k}) = 6 \hat{i}+2 \hat{j}-8 \hat{k}$
Next,calculate $\vec{a}-\vec{b}$:
$\vec{a}-\vec{b}=(5 \hat{i}-\hat{j}-3 \hat{k})-(\hat{i}+3 \hat{j}-5 \hat{k}) = 4 \hat{i}-4 \hat{j}+2 \hat{k}$
Now,find the dot product $(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b})$:
$(\vec{a}+\vec{b}) \cdot (\vec{a}-\vec{b}) = (6 \hat{i}+2 \hat{j}-8 \hat{k}) \cdot (4 \hat{i}-4 \hat{j}+2 \hat{k})$
$= (6)(4) + (2)(-4) + (-8)(2)$
$= 24 - 8 - 16$
$= 24 - 24 = 0$
Since the dot product is $0$,the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular.
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