Find a vector of magnitude 5units, and parallel to the resultant of the vectors \(\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\space and\space \vec{b}=\hat{i}-2\hat{j}+\hat{k}.\)
We have,
\(\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\space and\space \vec{b}=\hat{i}-2\hat{j}+\hat{k}.\)
Let \(\vec{c}\) be the resultant of a→and b→.
Then,
\(\vec{c}=\vec{a}+\vec{b}=(2+1)\hat{i}+(3-2)\hat{j}+(-1+1)\hat{k}=3\hat{i}+\hat{j}\)
\(∴|\vec{c}|=\sqrt{3^{2}+1^{2}}\sqrt{9+1}=\sqrt{10}\)
\(∴\hat{c}=\frac{\vec{c}}{|\vec{c}|}=\frac{(3\hat{i}+\hat{j})}{\sqrt{10}}\)
Hence,the vector of magnitude 5units and parallel to the resultant of vectors \(\vec{a}\) and \(\vec{b}\) is \(\pm5.\hat{c}=\)\(\pm5.\frac{1}{\sqrt{10}}(3\hat{i}+\hat{j})\)\(=\pm\frac{3\sqrt{10}\hat{i}}{2}\pm\frac{\sqrt{10}}{2}\hat{j}.\)
The area of a parallelogram whose diagonals are given by $ \vec{u} + \vec{v} $ and $ \vec{v} + \vec{w} $, where:
$ \vec{u} = 2\hat{i} - 3\hat{j} + \hat{k}, \quad \vec{v} = -\hat{i} + \hat{k}, \quad \vec{w} = 2\hat{j} - \hat{k} $ is:
The direction ratios of the normal to the plane passing through the points
$ (1, 2, -3), \quad (1, -2, 1) \quad \text{and parallel to the line} \quad \frac{x - 2}{2} = \frac{y + 1}{3} = \frac{z}{4} \text{ is:} $
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: