Question

If the vector components of a vector \(\vec a\) along a vector \[ \vec b=4\hat i+5\hat j+3\hat k \] and perpendicular to \(\vec b\) are respectively \[ \frac{7}{25}(4\hat i+5\hat j+3\hat k) \] and \[ \frac{1}{25}(47\hat i-10\hat j-46\hat k), \] then \[ |\vec a|^2= \]

• A. \(6\)
• B. \(9\)
• C. \(11\)
• D. \(17\)

Hint:

When a vector is expressed as the sum of components parallel and perpendicular to another vector, the components are orthogonal. Therefore, use \( |\vec a|^2=|\vec a_{\parallel}|^2+|\vec a_{\perp}|^2 \).

Answer 1

The Correct Option is C

Concept: If a vector is resolved into two perpendicular components, then by Pythagoras theorem, \[ |\vec a|^2 = |\vec a_{\parallel}|^2 + |\vec a_{\perp}|^2. \]

Step 1: Find the magnitude squared of the component along \(\vec b\). The component along \(\vec b\) is \[ \vec a_{\parallel} = \frac{7}{25}(4\hat i+5\hat j+3\hat k). \] Hence, \[ |\vec a_{\parallel}|^2 = \left(\frac{7}{25}\right)^2 (4^2+5^2+3^2). \] \[ = \frac{49}{625}(16+25+9). \] \[ = \frac{49}{625}\times 50 = \frac{98}{25}. \]

Step 2: Find the magnitude squared of the perpendicular component. The perpendicular component is \[ \vec a_{\perp} = \frac{1}{25}(47\hat i-10\hat j-46\hat k). \] Therefore, \[ |\vec a_{\perp}|^2 = \left(\frac{1}{25}\right)^2 (47^2+(-10)^2+(-46)^2). \] \[ = \frac{1}{625}(2209+100+2116). \] \[ = \frac{4425}{625} = \frac{177}{25}. \]

Step 3: Find \(|\vec a|^2\). Since the two components are perpendicular, \[ |\vec a|^2 = |\vec a_{\parallel}|^2 + |\vec a_{\perp}|^2. \] \[ = \frac{98}{25} + \frac{177}{25}. \] \[ = \frac{275}{25}. \] \[ =11. \] Therefore, \[ \boxed{|\vec a|^2=11} \] \[ \boxed{\text{Answer = (C)}} \]