Question

Find a vector which is parallel to v=î-2y and has a magnitude 10. ​

Answer 1

Answer:

Vector A is,

$A=xi+yj\\A=2\sqrt{5}i-4\sqrt{5}j$

Explanation:

In the question,

We have a vector,

v = i - 2j

We have to find a vector which is parallel to 'v' vector and has a magnitude of 10.

So,

Let us say the vector is A = xi + yj

So,

$|A| =\sqrt{x^{2}+y^{2}} \\10=\sqrt{x^{2}+y^{2}}\\100=x^{2}+y^{2}$

Now,

Also,

We know that in the vector, v, we have,

$tan\theta=\frac{y}{x}=\frac{-2}{1}\\So,\\y=-2x$

Now, on putting it in the equation we get,

$x^{2}+(-2x)^{2}=100\\x^{2}+4x^{2}=100\\5x^{2}=100\\x^{2}=20\\x=2\sqrt{5}$

So,

$y=-2x\\y=-2(2\sqrt{5})\\y=-4\sqrt{5}$

Therefore, the vector A is given by,

$A=xi+yj\\A=2\sqrt{5}i-4\sqrt{5}j$

Answer 2

The parallel vector will be$\left (\hat i \sqrt{5} -2\sqrt{5} \hat j \right ).$

Explanation:

This can be solved in the easiest manner.

Since, we know that;

• If two vectors $\vec A$and $\vec B$ are parallel to each other.

Then, we can represent them as

• $\vec A = n \vec B$                                          ........(1)

Therefore; we can write as;

• $\left | \vec A \right | = n \left | \vec B \right |$ .......(2)

• Where; n is a scalar quantity.

Now, it is given that;

• $\vec v = \hat i -2 \hat j$

       $\left | \vec v \right | = \sqrt{1^2 + (-2)^2} = \sqrt{5}$

It means, n =$\sqrt{5}$

• Now, the vector which is parallel to $\vec v$ will be, n$\vec v$.

• $n \ \vec v = \sqrt{5}\left (\hat i -2 \hat j\right )$

Thus, the parallel vector will be $\left (\hat i \sqrt{5} -2\sqrt{5} \hat j \right )$.