TS EAMCET · Maths · Vector Algebra
If \(M\) and \(N\) are the mid-points of the sides \(B C\) and \(C D\) respectively of a parallelogram \(A B C D\), then \(\mathbf{A M}+\mathbf{A N}\) equals
- A \(\frac{4}{3} \mathrm{AC}\)
- B \(\frac{5}{3} \mathrm{AC}\)
- C \(\frac{3}{2} \mathrm{AC}\)
- D \(\frac{6}{5} \mathrm{AC}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{2} \mathrm{AC}\)
Step-by-step Solution
Detailed explanation
\(\text {Since, } M \text { and } N \text { are the mid-points of } B C \text { and } C D \text {. }\) \(\begin{aligned} & \therefore & B M & =\frac{1}{2} \mathbf{b} \\ & \text {and } & D N & =\frac{1}{2} \mathbf{a} \end{aligned}\) Now, in \(\triangle A B M\),…
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