TS EAMCET · Maths · Vector Algebra
\(\mathbf{x}, \mathbf{y}, \mathbf{z}\) are three vectors each of magnitude \(\sqrt{2}\) and each making an angle \(60^{\circ}\) with one another. If \(\mathbf{a}=\mathbf{x} \times(\mathbf{y} \times \mathbf{z}), \mathbf{b}=\mathbf{y} \times(\mathbf{z} \times \mathbf{x})\), \(\mathbf{c}=\mathbf{x} \times \mathbf{y}\), then \(\mathbf{x}=\)
- A \(\frac{1}{2}[(a+b) \times c-(a+b)]\)
- B \(\frac{1}{2}[c+a-b]\)
- C \(\frac{1}{2}[(a+b) \times c+(a+b)]\)
- D \(\frac{1}{2}[(a \times b) \times c-a+b]\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{2}[(a+b) \times c-(a+b)]\)
Step-by-step Solution
Detailed explanation
Given that, \(|\mathbf{x}|=|\mathbf{y}|=|\mathbf{z}|=\sqrt{2}\) and \(\theta=60^{\circ}\) Thus, \(\mathbf{x} \cdot \mathbf{y}=|\mathbf{x}||\mathbf{y}| \cos \theta=(\sqrt{2})(\sqrt{2}) \cos 60^{\circ}\) \(=2 \times \frac{1}{2}=1\) Similarly…
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