Let \overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k and a vector \overrightarrow b be such that \overrightarrow a \times \overrightarrow b = \overrightarrow c and \overrightarrow a .\overrightarrow b = 3. Then | \overrightarrow b | equals :

Vector Algebra PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q43 with Solution

View written solutionFree

Correct answer: C

Given vectors $\vec a = \hat i + \hat j + \hat k = (1,1,1), \qquad \vec c = \hat j - \hat k = (0,1,-1).$

We are also given: $\vec a \times \vec b = \vec c, \qquad \vec a \cdot \vec b = 3.$
Use the identity relating dot product, cross product, and magnitude

For any vectors $\vec a, \vec b$ , $|\vec a \times \vec b|^2 + (\vec a \cdot \vec b)^2 = |\vec a|^2 |\vec b|^2.$

So we compute each quantity.
Compute $|\vec a|^2$ $|\vec a|^2 = 1^2 + 1^2 + 1^2 = 3.$
Compute $|\vec a \times \vec b|^2 = |\vec c|^2$ Since $\vec a \times \vec b = \vec c = (0,1,-1)$ , $|\vec c|^2 = 0^2 + 1^2 + (-1)^2 = 2.$
Use $\vec a \cdot \vec b = 3$ $(\vec a \cdot \vec b)^2 = 3^2 = 9.$
Substitute into the identity $|\vec a \times \vec b|^2 + (\vec a \cdot \vec b)^2 = |\vec a|^2 |\vec b|^2$ gives $2 + 9 = 3|\vec b|^2.$ $11 = 3|\vec b|^2.$ $|\vec b|^2 = \frac{11}{3}.$ Hence, $|\vec b| = \sqrt{\frac{11}{3}}.$
Match with options $\sqrt{\frac{11}{3}}$ corresponds to Option C.

Final Answer: Option C, $\boxed{\sqrt{\frac{11}{3}}}$