Two vectors `vec(P) " and "vec(Q)` are added, the magnitude of resultant is 15 units. If `vec(Q)` is reversed and added to `vec(P)` resultant has a magnitude `sqrt(113)` units. The resultant of `vec(P)` and a vector perpendicular ` vec(P)` and equal in magnitude to` vec(Q)` has a magnitude

Options

A. 13 units
B. 17 units
C. 19 units
D. 20 units

Explanation

To solve the problem step by step, we will follow the reasoning outlined in the video transcript.

### Step 1: Establish the first equation
We know that when vectors \( \vec{P} \) and \( \vec{Q} \) are added, the magnitude of the resultant vector \( \vec{R_1} \) is 15 units. We can write this as:
\[ |\vec{R_1}| = |\vec{P} + \vec{Q}| = 15 \]
Using the formula for the magnitude of the sum of two vectors:
\[ |\vec{R_1}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta} \]
Squaring both sides gives us:
\[ |\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta = 15^2 = 225 \]
This is our **Equation 1**.

### Step 2: Establish the second equation
Next, when \( \vec{Q} \) is reversed and added to \( \vec{P} \), the magnitude of the resultant vector \( \vec{R_2} \) is \( \sqrt{113} \) units. We can express this as:
\[ |\vec{R_2}| = |\vec{P} - \vec{Q}| = \sqrt{113} \]
Using the formula for the magnitude of the difference of two vectors:
\[ |\vec{R_2}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2 - 2 |\vec{P}| |\vec{Q}| \cos \theta} \]
Squaring both sides gives us:
\[ |\vec{P}|^2 + |\vec{Q}|^2 - 2 |\vec{P}| |\vec{Q}| \cos \theta = 113 \]
This is our **Equation 2**.

### Step 3: Add the two equations
Now, we will add Equation 1 and Equation 2:
\[ (|\vec{P}|^2 + |\vec{Q}|^2 + 2 |\vec{P}| |\vec{Q}| \cos \theta) + (|\vec{P}|^2 + |\vec{Q}|^2 - 2 |\vec{P}| |\vec{Q}| \cos \theta) = 225 + 113 \]
This simplifies to:
\[ 2 |\vec{P}|^2 + 2 |\vec{Q}|^2 = 338 \]
Dividing by 2 gives:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 169 \]

### Step 4: Find the magnitude of the resultant vector
Now, we need to find the magnitude of the resultant vector when \( \vec{P} \) is added to a vector \( \vec{B} \) that is perpendicular to \( \vec{P} \) and equal in magnitude to \( \vec{Q} \).

Since \( \vec{B} \) is perpendicular to \( \vec{P} \), we can express the magnitude of the resultant vector \( \vec{R_3} \) as:
\[ |\vec{R_3}| = |\vec{P} + \vec{B}| = \sqrt{|\vec{P}|^2 + |\vec{B}|^2} \]
Given that \( |\vec{B}| = |\vec{Q}| \), we can substitute:
\[ |\vec{R_3}| = \sqrt{|\vec{P}|^2 + |\vec{Q}|^2} \]
From our earlier calculation, we know:
\[ |\vec{P}|^2 + |\vec{Q}|^2 = 169 \]
Thus,
\[ |\vec{R_3}| = \sqrt{169} = 13 \]

### Final Answer
The magnitude of the resultant vector \( |\vec{R_3}| \) is **13 units**.

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