Question

a vector perpendicular to the vector (i+2j) and having magnitude 3√5 units is

Answer 1

Hope you get the answer

Answer 2

"The vectors perpendicular to the vector i + 2 j with magnitude $3 \sqrt{5}$are (6i - 3j) or r = (6i + 3j)

Solution:

We can derive the vector from the vector dot product formula.

The vector r be xi + yj and perpen`dicular to the vector given.

Thereby, the dot product between two perpendicular vectors should be zero.

r.a = (x i + y j).( i + 2 j)

r.a = x + 2 y = 0

x = -2y

So, for the magnitude we can use the formula given below,

$r=\sqrt{x^{2}+y^{2}}=3 \sqrt{5}$

Squaring both the sides, we get,

$x^{2}+y^{2}=45$

On substituting value of x, we get,

$(-2 y)^{2}+y^{2}=45$

$5 y^{2}=45$

$y^{2}=9$

$y=\pm 3$

$\Rightarrow x=-2 y=-2(3)=-6$

$\Rightarrow x=2 y=-2(-3)=6$

Therefore,

r = (6i - 3j) or r = (6i + 3j)"