Question

A vector A which has magnitude 8.0 is added to a vector B which lies on x-axis.The sum of these two vectors lies on y-axis and has a magnitude twice of the magnitude of vector B.The magnitude of vector B is
a) 8
b) 1.41*8
c) 8/(5)^1/2
d) 8*(5)^1/2

Answer 1

1. A vector $A$ which has magnitude 8.0 is added to a vector $B$ which lies on $x$-axis. The sum of these two vectors lies on $y$-axis and has a magnitude twice of the magnitude of vector $B$. The magnitude of vector $B$ is:

a) 8;

b) $1.41 \times 8$;

c) $8 / (5)^{1 / 2}$;

d) $8 \times (5)^{1 / 2}$.

Solution.

As vector $B$ which lies on $x$-axis, then $\vec{B}(b;0)$. The magnitude of vector $B$ is $\left|\vec{B}\right| = \sqrt{b^2 + 0^2} = |b|$.

Assume that $\vec{A}(a_1; a_2)$.

The sum of these two vectors is $\vec{C} = \vec{A} + \vec{B} = (a_1; a_2) + (b; 0) = (a_1 + b; a_2)$. This vector lies on $y$-axis, so $a_1 + b = 0$, $a_1 = -b$.

The magnitude of vector $\vec{A}$ is $\left|\vec{A}\right| = \sqrt{a_1^2 + a_2^2} = \left| (-b)^2 + a_2^2 \right| = \sqrt{b^2 + a_2^2}$. According to the condition of the problem, $\left|\vec{A}\right| = 8.0$.

The magnitude of vector $\vec{C}$ is $\left|\vec{C}\right| = \sqrt{0^2 + a_2^2} = |a_2|$. According to the condition of the problem, $\left|\vec{C}\right| = 2 \cdot \left|\vec{B}\right|$, so $|a_2| = 2 \cdot |b|$.

Summarizing, $\left|\vec{A}\right| = \sqrt{b^2 + (2|b|)^2} = 8.0$, $|b| = \frac{8.0}{\sqrt{1 + 2^2}} = \frac{8.0}{\sqrt{5}}$.

Thus, the magnitude of vector $\vec{B}$ is $\left|\vec{B}\right| = |b| = \frac{8.0}{\sqrt{5}}$.

Answer: $C$.