Question

The resultant of two vectors is perpendicular to one
of them and has the magnitude 4 m. If the sum of
the magnitude of two vectors is 8 m then their
respective magnitude are
(1) 4 m, 4 m
(2) 2 m, 6 m
(3) 3 m, 5 m
(4) 1 m, 7 m​

Answer 1

Let us assume that the two vectors are M and N & resultant vector of M and N is J.

Given that, the resultant of two vectors is perpendicular to one of them and has the magnitude 4 m (J = 4 m).Also, the sum of the magnitude of two vectors is 8 m (M + N = 8 m)

We have to find the respective magnitude of the assumed vectors.

[ Refer the attachment for figure. Triangle law applied on it.

It states that if two vectors working together at a point are represented by two sides of a triangle taken in the same order in magnitude and direction. On the other hand, their resultant vector is shown by the third side of the triangle taken in the opposite order. ]

Using Pythagorean theorem,

→ (J)² + (M)² = (N)²

Substitute the known values,

→ (4)² + (M)² = (N)²

[ M + N = 8, M = 8 - N ]

→ 16 + (8 - N)² = N²

Using identity: (a-b)² = a² - 2ab + b²

→ 16 + 64 - 16N + N² = N²

→ 80 - 16N = N² - N²

→ 80 = 16N

→ N = 80/16

→ N = 5

Substitute value of N = 5 in M + N = 8

→ M + 5 = 8

→ M = 8 - 5

→ M = 3

Therefore, their respective magnitudes are 3m and 5 m.

Option (3) 3m, 5 m

Answer 2

GIVEN :-

◙ The resultant of two vectors is perpendicular to one  of them, and has the magnitude of 4 m.

◙ The sum of the magnitude of two vectors is 8 m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

TO FIND :-

The magnitude of the vectors out of :-

(1) 4 m, 4 m

(2) 2 m, 6 m

(3) 3 m, 5 m

(4) 1 m, 7 m​

SOLUTION :-

Let the two vectors be $\sf{\overrightarrow{a}}$ and $\sf{\overrightarrow{b}}$. Let their resultant vector be $\sf{\overrightarrow{R}}$.

$\blacksquare$ $\sf{\overrightarrow{R}=4\ m}$   (given)

$\blacksquare\ \sf{|\overrightarrow{a}|+|\overrightarrow{b}|=8\ m}$   (given)

$\sf{\longrightarrow |\overrightarrow{b}|=8-|\overrightarrow{a}|}$

It is also given that, the angle between the angles is θ = 90°. Hence, we obtain a right triangle.

Applying Pythagoras Theorem :-

$\sf{\overrightarrow{R}^2+(|\overrightarrow{a}|)^2=(|\overrightarrow{b}|)^2}$

$\sf{\implies 4^2+(|\overrightarrow{a}|)^2=(8-|\overrightarrow{a}|)^2}\\\\\sf{\implies 16+(|\overrightarrow{a}|)^2=8^2+(|\overrightarrow{a}|)^2-2\times8\times |\overrightarrow{a}|}\\\\\sf{\implies 16=64-16|\overrightarrow{a}|}\\\\\sf{\implies 16 |\overrightarrow{a}|=48}\\\\\boxed{\sf{\implies |\overrightarrow{a}|=3\ m}}$

$\rule{150}2$

We know,

$\sf{\longrightarrow |\overrightarrow{b}|=8-|\overrightarrow{a}|}$

Substituting the value of $|\overrightarrow{a}|$ :-

$\sf{\Longrightarrow |\overrightarrow{b}|=8-3}\\\\\boxed{\sf{\sf{\Longrightarrow |\overrightarrow{b}|=5\ m}}}$

Therefore, the answer is (3) 3 m, 5 m.