Question

If \[ \vec{a}=4\vec{i}+6\vec{j},\quad \vec{b}=3\vec{j}+4\vec{k} \] and \(\vec{c}\) is the projection vector of \(\vec{a}\) on \(\vec{b}\), then \(\vec{c}\) and \(|\vec{c}|\) respectively are:

• A. \(\dfrac{18}{25}\vec{b},\ \dfrac{18}{5}\)
• B. \(\dfrac{18}{5}\vec{b},\ 18\)
• C. \(\dfrac{25}{18}\vec{b},\ \dfrac{18}{5}\)
• D. \(\dfrac{5}{18}\vec{b},\ \dfrac{5}{18}\)

Hint:

The projection vector of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}\), while its magnitude is \(\frac{|\vec{a}\cdot\vec{b}|}{|\vec{b}|}\).

Answer 1

The Correct Option is A

Step 1: Use the formula for projection vector.
The projection vector of \(\vec{a}\) on \(\vec{b}\) is \[ \vec{c}=\frac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\vec{b}. \]

Step 2: Calculate \(\vec{a}\cdot\vec{b}\).
Given, \[ \vec{a}=4\vec{i}+6\vec{j} \] and \[ \vec{b}=3\vec{j}+4\vec{k}. \] Therefore, \[ \vec{a}\cdot\vec{b} = (4)(0)+(6)(3)+(0)(4). \] \[ \vec{a}\cdot\vec{b}=18. \]

Step 3: Calculate \(|\vec{b}|^2\).
\[ |\vec{b}|^2=0^2+3^2+4^2. \] \[ |\vec{b}|^2=9+16=25. \]

Step 4: Find the projection vector.
\[ \vec{c} = \frac{18}{25}\vec{b}. \]

Step 5: Find \(|\vec{c}|\).
Since \[ |\vec{b}|=\sqrt{25}=5, \] we get \[ |\vec{c}| = \left|\frac{18}{25}\vec{b}\right|. \] \[ |\vec{c}| = \frac{18}{25}|\vec{b}|. \] \[ |\vec{c}| = \frac{18}{25}\times 5. \] \[ |\vec{c}|=\frac{18}{5}. \]

Step 6: Final conclusion.
Therefore, \[ \boxed{\vec{c}=\frac{18}{25}\vec{b},\quad |\vec{c}|=\frac{18}{5}} \]