TS EAMCET · Maths · Vector Algebra
If \(\mathbf{a}\) and \(\mathbf{b}\) are two non-zero perpendicular vectors, then a vector \(\mathbf{y}\) satisfying equations \(\mathbf{a} \cdot \mathbf{y}=c\) (where, \(c\) is scalar) and \(\mathbf{a} \times \mathbf{y}=\mathbf{b}\) is
- A \(|\mathbf{a}|^2[\mathrm{c} \mathbf{a}-(\mathbf{a} \times \mathbf{b})]\)
- B \(|\mathbf{a}|^2 \cdot[\mathbf{c} \mathbf{a}+(\mathbf{a} \times \mathbf{b})]\)
- C \(\frac{1}{|\mathbf{a}|^2}[c \mathbf{a}-(\mathbf{a} \times \mathbf{b})]\)
- D \(\frac{1}{|\mathbf{a}|^2}[c \mathbf{a}+(\mathbf{a} \times \mathbf{b})]\)
Answer & Solution
Correct Answer
(C) \(\frac{1}{|\mathbf{a}|^2}[c \mathbf{a}-(\mathbf{a} \times \mathbf{b})]\)
Step-by-step Solution
Detailed explanation
Since, \(\mathbf{a}, \mathbf{b}\) and \(\mathbf{a} \times \mathbf{b}\) are non-coplanar, hence \(\mathbf{y}=\mathbf{a} x+t b \mathbf{x}+(\mathbf{a} \times \mathbf{b}) z\) For some scalars \(x, t\) and \(z\), Now,…
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