Find a vector which is parallel to v =i-2 j and has a magnitude 1…

Question

Find a vector which is parallel to $\overrightarrow{ v }=\hat{i}-2 \hat{j}$ and has a magnitude $10.$

✓

Answer

Let the vector be $\vec{w}=w_x \hat{i}+w_y \hat{j}$
$ |\vec{w}|=\sqrt{w_x^2+w_y^2}=10$
$\therefore \quad w_x^2+w_y^2=100$
Also, $v \cdot W = VW$
$\ldots(\because|\vec{v}|$ and $|\vec{w}|$ are parallel vectors)
$\Rightarrow(\hat{ i }-2 \hat{ j }) \cdot\left( w _{ x } \hat{ i }+ w _y \hat{ j }\right)=\sqrt{(1)^2+(-2)^2} \times 10$
$\ldots\left(\because|\overrightarrow{ v }|=\sqrt{(1)^2+(-2)^2}\right)$
$\therefore \quad w _{ x }-2 w _{ y }=10 \sqrt{5} \quad \ldots \text { (ii) }$
Substituting for $w_x$ in $(i)$ using equation $(ii),$
$\left(10 \sqrt{5}+2 w_y\right)^2+w_y^2=100$
$\therefore \quad 500+40 \sqrt{5} w_y+4 w_y^2-100=0$
$\therefore \quad 5 w_y^2+40 \sqrt{5} w_y+400=0$
$\therefore \quad w_y^2+8 \sqrt{5} w_y+80=0$
Using factorisation formula,
$x=\frac{-b \pm \sqrt{b^2-4 a c}}{2 a}$
$w_y=\frac{-8 \sqrt{5} \pm \sqrt{(8 \sqrt{5})^2-4 \times 1 \times 80}}{2 \times 1}$
$=w_y=\frac{-8 \sqrt{5} \pm 0}{2}=-4 \sqrt{5}=\frac{-20}{\sqrt{5}}$
Using equation $(ii),$
$w_x =10 \sqrt{5}+2\left(\frac{-20}{\sqrt{5}}\right)$
$ =10 \sqrt{5}-\frac{40}{\sqrt{5}}$
$ =\frac{(10 \sqrt{5} \times \sqrt{5})-40}{\sqrt{5}}=\frac{50-40}{\sqrt{5}}=\frac{10}{\sqrt{5}}$
$\therefore \quad \vec{w} =w_x \hat{i}+w_y \hat{j}=\frac{10}{\sqrt{5}} \hat{i}-\frac{20}{\sqrt{5}} \hat{j}$
Answer:
Required vector is $\frac{10}{\sqrt{5}} \hat{ i }-\frac{20}{\sqrt{5}} \hat{ j }$
Alternate method:
When two vectors are parallel, one vector is scalar multiple of another, i.e., if $\vec{v}$ and $\vec{w}$ are parallel then, $\vec{w}=n \vec{v}$ where, $n$ is scalar.
This means, $\vec{w}=n v_x \hat{i}+n v_y \hat{j}$
$=n \hat{i}-2 \hat{j} \quad \quad \ldots .\left(\because v_x=1, v_y=2\right)$
$\therefore \quad|\vec{w}|=\sqrt{(n)^2+(-2 n)^2}=\sqrt{5} n$
Given: $|\vec{w}|=10$
$\therefore \quad n=\frac{10}{\sqrt{5}}=2 \sqrt{5}$
$\therefore \quad \overrightarrow{ w }=2 \sqrt{5} \hat{ i }-2(2 \sqrt{5}) \hat{ j }$
$=2 \sqrt{5} \hat{ i }-4 \sqrt{5} \hat{ j }$
$=\frac{2 \sqrt{5} \times \sqrt{5}}{\sqrt{5}} \hat{ i }-\frac{4 \sqrt{5} \times \sqrt{5}}{\sqrt{5}} \hat{ j }$
$\therefore \quad \vec{w}=\frac{10}{\sqrt{5}} \hat{ i }-\frac{20}{\sqrt{5}} \hat{ j }$