Question
Download Solution PDFIf \(\rm \vec{a}\) and \(\rm \vec{b}\) are mutually perpendicular unit vectors, then \(\rm (3\vec{a}+2\vec{b})\cdot (5\vec{a}-6\vec{b})=\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFConcept:
If \(\rm \vec{a}\) and \(\rm \vec{b}\) are mutually perpendicular unit vectors \(\rm \vec{a}.\vec{b} = 0= \)\(\rm \vec{b}.\vec{a} = 0\)
\(\rm \vec{a}.\vec{a} = |\vec{a}|^2\), \(\rm \vec{b}.\vec{b} = |\vec{b}|^2\)
Calculations:
Consider, \(\rm (3\vec{a}+2\vec{b})\cdot (5\vec{a}-6\vec{b})\)
Given \(\rm \vec{a}\) and \(\rm \vec{b}\) are mutually perpendicular unit vectors.
So, \(\rm \vec{a}.\vec{b} = 0= \)\(\rm \vec{b}.\vec{a} = 0\)
And \(\rm \vec{a}\) and \(\rm \vec{b}\) are unit vectors
So, \(\rm |\vec{a}|= |\vec{b}| = 1\)
\(\rm (3\vec{a}+2\vec{b})\cdot (5\vec{a}-6\vec{b})\)
\(= \rm 3\vec{a}\cdot 5\vec{a}+2\vec{b}\cdot 5\vec{a} -3\vec{a}\cdot6\vec{b}-2\vec{b}\cdot 6\vec{b}\)
= \(\rm 15|\vec {a}|^2 - 12 |\vec {b}|^2\)
= 15 - 12
= 3
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