Let \( \vec{a},\vec{b},\vec{c} \) are three vectors such that \( |\vec{a}|=3,|\vec{b}|=2,|\vec{c}|=1 \). If the projection of \( \vec{b} \) along \( \vec{c} \) is equal to the projection of \( \vec{a} \) along \( \vec{c} \) and \( \vec{b}\perp\vec{a} \), then the value of \( |2\vec{a}-2\vec{b}+3\vec{c}| \) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \sqrt{61} \)
Both conditions are statements about dot products. The projection of \( \vec{b} \) along \( \vec{c} \) is \( \frac{\vec{b}\cdot\vec{c}}{|\vec{c}|} \), and similarly for \( \vec{a} \).
\( \displaystyle \frac{\vec{b}\cdot\vec{c}}{|\vec{c}|}=\frac{\vec{a}\cdot\vec{c}}{|\vec{c}|}\ \Rightarrow\ \vec{a}\cdot\vec{c}=\vec{b}\cdot\vec{c} \)
\( \displaystyle \vec{b}\perp\vec{a}\ \Rightarrow\ \vec{a}\cdot\vec{b}=0 \)
Now we square the required magnitude, using \( |\vec{u}|^2=\vec{u}\cdot\vec{u} \):
\( \displaystyle |2\vec{a}-2\vec{b}+3\vec{c}|^2=4|\vec{a}|^2+4|\vec{b}|^2+9|\vec{c}|^2-8\,\vec{a}\cdot\vec{b}+12\,\vec{a}\cdot\vec{c}-12\,\vec{b}\cdot\vec{c} \)
\( \displaystyle =4(9)+4(4)+9(1)-8(0)+12(\vec{a}\cdot\vec{c}-\vec{b}\cdot\vec{c}) \)
\( \displaystyle =36+16+9+0+0 \)
\( \displaystyle =61 \)
\( \displaystyle |2\vec{a}-2\vec{b}+3\vec{c}|=\sqrt{61} \)