91Ó°ÊÓ

What should be added in vector \(\mathbf{a}=3 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) to get its resultant a unit vector \(\hat{\mathbf{i}}\) ? (a) \(-2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\) (b) \(-2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) (c) \(2 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\) (d) None of these

The position vector of the points which divides internally in the ratio \(2: 3\) the join of the points \(2 \mathrm{a}-3 \mathrm{~b}\) and \(3 \mathrm{a}-2 \mathbf{b}\), is (a) \(\frac{12}{5} a+\frac{13}{5} b\) (b) \(\frac{12}{5}=\frac{13}{5} b\) (c) \(\frac{3}{5} a-\frac{2}{5} b\) (d) None of these

If \(O\) is origin and \(C\) is the mid-point of \(A(2,-1)\) and \(B(-4,3)\) Then, value of \(O C\) is (a) \(\hat{i}+j\) (b) \(i-\hat{j}\) (c) \(-\mathbf{i}+\mathbf{j}\) (d) \(-\hat{i}-\hat{j}\)

If \(\mathrm{a}=(1,-1)\) and \(\mathbf{b}=(-2, \mathrm{~m})\) are two collinear vectors, then \(m\) is equal to (a) 4 (b) 3 (c) 2 (d) 0

In the \(\triangle O A B, M\) is the mid-point of \(A B, C\) is a point on \(O M\), such that \(2 \mathrm{OC}=\mathbf{C M} \cdot X\) is a point on the side \(O B\) such that \(\mathrm{OX}=2 \mathrm{XB}\). The line \(X \mathrm{C}\) is produced to meet \(O A\) in \(Y\). Then, \(\frac{O Y}{Y A}\) is equal to (a) \(\frac{1}{3}\) (b) \(\frac{2}{7}\) (c) \(\frac{3}{2}\) (d) \(\frac{2}{5}\)

Let \(\mathrm{p}\) be the position vector of orthocentre and \(\mathrm{g}\) is the position vector of the centroid of \(\triangle A B C\), where circumcentre is the origin. If \(\mathbf{p}=k g\), then the value of \(k\) is