JEE Mains · Maths · STD 12 - 10. vector algebra
Let \(\vec a = \hat i - \hat j,\) \(\vec b = \hat i + \hat j + \hat k\) and \(\vec c\) be a vector such that \(\vec a \times \vec c + \vec b = 0\) and \(\vec a.\vec c = 4\), then \({\left| {\vec c} \right|^2}\) is equal to
- A \(\frac{{19}}{2}\)
- B \(9\)
- C \(8\)
- D \(\frac{{17}}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{{19}}{2}\)
Step-by-step Solution
Detailed explanation
\(a=\hat{i}-\hat{j}, b=\hat{i}+\hat{j}+\hat{k}, c=x \hat{i}+y \hat{j}+z \hat{k}\) \(\vec{a} \times \vec{c}+\vec{b}=0\) \( \Rightarrow \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 1&{ - 1}&0\\ x&y&z \end{array}} \right|\) \(+(\hat{i}+\hat{\bar{j}}+\hat{k})=0\)…
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