Vectors 4 Question 15
15. (i) If $C$ is a given non-zero scalar and $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{B}}$ be given non-zero vectors such that $\overrightarrow{\mathbf{A}} \perp \overrightarrow{\mathbf{B}}$, then find the vector $\overrightarrow{\mathbf{X}}$ which satisfies the equations $\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{X}}=c$ and $\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{X}}=\overrightarrow{\mathbf{B}}$
(ii) $\overrightarrow{\mathbf{A}}$ vector $A$ has components $A _1, A _2, A _3$ in a right-handed rectangular cartesian coordinate system OXYZ. The coordinate system is rotated about the $X$-axis through an angle $\frac{\pi}{2}$. Find the components of $A$ in the new coordinate system, in terms of $A _1, A _2, A _3$.
(1983, 2M)
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Solution:
- (i) Given, $\overrightarrow{\mathbf{A}} \perp \overrightarrow{\mathbf{B}} \Rightarrow \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}=0$
and $\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{X}}=\overrightarrow{\mathbf{B}} \Rightarrow \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}=0$ and $\overrightarrow{\mathbf{X}} \cdot \overrightarrow{\mathbf{B}}=0$
Now, $\quad[\overrightarrow{\mathbf{X}} \overrightarrow{\mathbf{A}} \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}]=\overrightarrow{\mathbf{X}} \cdot{\overrightarrow{\mathbf{A}} \times(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}})}$
$$ =\overrightarrow{\mathbf{X}} \cdot{(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}}) \overrightarrow{\mathbf{A}}-(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{A}}) \overrightarrow{\mathbf{B}}} $$
$$ =(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}})(\overrightarrow{\mathbf{X}} \cdot \overrightarrow{\mathbf{A}})-(\overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{A}})(\overrightarrow{\mathbf{X}} \cdot \overrightarrow{\mathbf{B}})=0 $$
$\Rightarrow \overrightarrow{\mathbf{X}}, \overrightarrow{\mathbf{A}}, \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}$ are coplanar.
So, $\overrightarrow{\mathbf{X}}$ can be represented as a linear combination of $\overrightarrow{\mathbf{A}}$ and $\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}}$. Let us consider, $\quad \overrightarrow{\mathbf{X}}=l \overrightarrow{\mathbf{A}}+m(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}})$
Since,
$$ \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{X}}=c $$
$\therefore \overrightarrow{\mathbf{A}} \cdot{l \overrightarrow{\mathbf{A}}+m(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}})}=c$
$\Rightarrow \quad l|\overrightarrow{\mathbf{A}}|^{2}+0=c$
$\Rightarrow \quad l=\frac{c}{|\overrightarrow{\mathbf{A}}|^{2}}$
Also, $\quad \overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{X}}=\overrightarrow{\mathbf{B}} \Rightarrow \overrightarrow{\mathbf{A}} \times{l \overrightarrow{\mathbf{A}}+m(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}})}=\overrightarrow{\mathbf{B}}$
$\Rightarrow \quad l(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{A}})+m{\overrightarrow{\mathbf{A}} \times(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}})}=\overrightarrow{\mathbf{B}}$
$\Rightarrow \quad 0-m|\overrightarrow{\mathbf{A}}|^{2} \overrightarrow{\mathbf{B}}=\overrightarrow{\mathbf{B}}$
$\Rightarrow \quad m=-\frac{1}{|\overrightarrow{\mathbf{A}}|^{2}}$
$\therefore \overrightarrow{\mathbf{X}}=\left(\frac{c}{|\overrightarrow{\mathbf{A}}|^{2}}\right) \overrightarrow{\mathbf{A}}-\left(\frac{1}{|\overrightarrow{\mathbf{A}}|^{2}}\right)(\overrightarrow{\mathbf{A}} \times \overrightarrow{\mathbf{B}})$
(ii) Since, vector $\overrightarrow{\mathbf{A}}$ has components $A _1, A _2, A _3$ in the coordinate system $O X Y Z$.
$\therefore \quad \overrightarrow{\mathbf{A}}=A _1 \hat{\mathbf{i}}+A _2 \hat{\mathbf{j}}+A _3 \hat{\mathbf{k}}$
When the given system is rotated about an angle of $\pi / 2$, the new $X$-axis is along old $Y$-axis and new $Y$-axis is along the old negative $X$-axis , whereas $z$ remains same.
Hence, the components of $A$ in the new system are $\left(A _2,-A _1, A _3\right)$.
$\therefore \quad \overrightarrow{\mathbf{A}}$ becomes $\left(A _2 \hat{\mathbf{i}}-A _1 \hat{\mathbf{j}}+A _3 \hat{\mathbf{k}}\right)$.
