Solveeit Logo

Question

Question: Two coinitial vectors \[\overrightarrow{A}\And \overrightarrow{B}\] terminate at a square plane of s...

Two coinitial vectors A&B\overrightarrow{A}\And \overrightarrow{B} terminate at a square plane of slide 323\sqrt{2}m. A\overrightarrow{A} is perpendicular to the square plane and terminates at the center of the square while B\overrightarrow{B} terminates at the corner of the square If the initial point of A&B\overrightarrow{A}\And \overrightarrow{B} is at distance of 4m. from the center of square plane then the ratio of magnitude of vectors
A&B\overrightarrow{A}\And \overrightarrow{B} is?
A. 3:4
B. 4:5
C. 5:6
D. 2:3

Explanation

Solution

The vectors A and B are coinitial. Coinitial vectors are those vectors which have the same starting point. Now, the two vectors terminate at the given square. The first vector A is perpendicular to the square and the vector B is not perpendicular to the square but at point B. So, there must be some angle between the two vectors.

Complete step by step answer:
Given, the initial point of A&B\overrightarrow{A}\And \overrightarrow{B}is at distance of 4m. from the center of square plane,
A=4m\left| A \right|=4m
If the side of the square is a then the length of the diagonal will be 2a\sqrt{2}a. Also, the diagonals bisect each other at right angles, so half of the length of the diagonal is a2\dfrac{a}{\sqrt{2}}. Now here in this problem, the length of the side of the square is 323\sqrt{2} m, so half of the diagonal length will be 3m. The length of the vector B\overrightarrow{B} can be find out by using Pythagoras theorem:

Let us assume the vector B meets the point D of the square and the centre of the square is O and the initial point of starting of the two vectors is P, then
\Rightarrow P{{D}^{2}}=A{{O}^{2}}+O{{D}^{2}} \\\ \Rightarrow P{{D}^{2}}={{4}^{2}}+{{3}^{2}} \\\ \Rightarrow PD=\sqrt{16+9} \\\ \Rightarrow PD=5 \\\
So, the magnitude of vector B is 5 m. Now taking the ratio of the magnitude of the vectors comes out to be
A:B=4:5\therefore \left| A \right|:\left| B \right|=4:5

So, the correct option is B.

Note: Most of the time we get confused that if the magnitude is zero then the given quantity cannot be termed as a vector but in actual it depends upon the state of the quantity. Here we just need to find the magnitude of the two vectors and take the ration. First vector A was perpendicular and the distance between the centre and the starting point of the vectors equals the magnitude of vector A.