TS EAMCET · Maths · Vector Algebra
If the position vectors of the points \(A, B, C, D\) given by \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\), \(\frac{1}{4}(7 \hat{\mathbf{i}}+15 \hat{\mathbf{j}}+15 \hat{\mathbf{k}}) \text { and } \frac{1}{3}[7 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+(5+3 a) \hat{\mathbf{k}}]\) respectively are such that \(|\mathbf{A C}|=|\mathbf{B D}|\), then \(16(3 a-1)^2=\)
- A 143
- B 139
- C 189
- D 187
Answer & Solution
Correct Answer
(D) 187
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