MHT CET · Maths · Vector Algebra
Let \(\bar{a}=3 \hat{i}+2 \hat{j}+x \hat{k}\) and \(\bar{b}=\hat{i}-\hat{j}+\widehat{k}\), for some real \(x\). Then \(|\bar{a} \times \bar{b}|=r\) is possible, if
- A \(0 < r \leq \sqrt{\frac{3}{2}}\)
- B \(r \geq 5 \sqrt{\frac{3}{2}}\)
- C \(3 \sqrt{\frac{3}{2}}< r<5 \sqrt{\frac{3}{2}}\)
- D \(\sqrt{\frac{3}{2}} \leq r \leq 3 \sqrt{\frac{3}{2}}\)
Answer & Solution
Correct Answer
(B) \(r \geq 5 \sqrt{\frac{3}{2}}\)
Step-by-step Solution
Detailed explanation
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1\end{array}\right|=(2+x) \hat{i}+(x-3) \hat{j}+(-3-2) \hat{k}\)
now \(r=|\vec{a} \times \vec{b}|=\sqrt{(2+x)^2+(x-3)^2+(-5)^2}\)
\(=\sqrt{2 x^2-2 x+38}\)
\(=\sqrt{2\left(x-\frac{1}{2}\right)^2+\frac{75}{2}}\)
\(\Rightarrow r_{\min }=\sqrt{\frac{75}{2}}\) at \(x=\frac{1}{2}\)
\(\Rightarrow r \geq 5 \cdot \sqrt{\frac{3}{2}}\)
now \(r=|\vec{a} \times \vec{b}|=\sqrt{(2+x)^2+(x-3)^2+(-5)^2}\)
\(=\sqrt{2 x^2-2 x+38}\)
\(=\sqrt{2\left(x-\frac{1}{2}\right)^2+\frac{75}{2}}\)
\(\Rightarrow r_{\min }=\sqrt{\frac{75}{2}}\) at \(x=\frac{1}{2}\)
\(\Rightarrow r \geq 5 \cdot \sqrt{\frac{3}{2}}\)
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