Question

If \(\vec{A}=\vec{B}+\vec{C}\) and magnitudes of \(\vec{A},\vec{B}\) and \(\vec{C}\) are 5, 4 and 3 units respectively. The angle between \(\vec{A}\) and \(\vec{C}\) is

• A. \(\left(\frac{\pi }{2}\right)^{c}\)
• B. \(\cos ^{-1}(4/5)\)
• C. \(\sin ^{-1}(3/4)\)
• D. \(\cos ^{-1}(3/5)\)

Hint:

Write the second vector as the difference of the other two and square it, then use the dot product to get the cosine of the angle.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
Since \(\vec{A}=\vec{B}+\vec{C}\), we can write \(\vec{B}=\vec{A}-\vec{C}\). Squaring this relates the magnitudes to the angle \(\theta\) between \(\vec{A}\) and \(\vec{C}\).

Step 2: Calculation
\(B^{2}=A^{2}+C^{2}-2AC\cos\theta\)
\(16=25+9-2(5)(3)\cos\theta\)

Step 3: Solving for the angle
\(30\cos\theta=18\), so \(\cos\theta=\dfrac{3}{5}\).

Step 4: Check
The sides 3, 4, 5 form a right triangle, and the side 3 is adjacent to the angle with hypotenuse 5, so \(\cos\theta=\dfrac{3}{5}\) again.

Final Answer:
The angle between the two vectors is \(\cos^{-1}(3/5)\). \[ \boxed{\text{(D) }\cos^{-1}\left(\dfrac{3}{5}\right)} \]