CUET Mathematics 2024 - If a, b and c are three vectors such that a+b+c=0, where a and b are unit vectors and |c|=2, then the angle between the vectors b and c is : | PYQs + Solutions | AfterBoards
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CUET Mathematics 2024 PYQs

CUET Mathematics 2024

Algebra
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Vector Algebra

Medium

If a,b\vec{a}, \vec{b} and c\vec{c} are three vectors such that a+b+c=0\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}, where a\vec{a} and b\vec{b} are unit vectors and c=2|\vec{c}|=2, then the angle between the vectors b\vec{b} and c\vec{c} is :

Correct Option: 4
Given: \newline - a+b+c=0\vec{a} + \vec{b} + \vec{c} = \vec{0} \newline - a=b=1|\vec{a}| = |\vec{b}| = 1 (unit vectors) \newline - c=2|\vec{c}| = 2
Rearranging the equation: c=ab\vec{c} = -\vec{a} - \vec{b}

The angle between vectors can be found using the dot product formula: \newline bc=bccosθ\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta
Where θ\theta is the angle between the vectors.

Calculating bc\vec{b} \cdot \vec{c}: \newline bc=b(ab)=babb\vec{b} \cdot \vec{c} = \vec{b} \cdot (-\vec{a} - \vec{b}) = -\vec{b} \cdot \vec{a} - \vec{b} \cdot \vec{b}
Since b\vec{b} is a unit vector: bb=b2=1\vec{b} \cdot \vec{b} = |\vec{b}|^2 = 1
So: bc=ba1\vec{b} \cdot \vec{c} = -\vec{b} \cdot \vec{a} - 1

Using the magnitude of c\vec{c}: \newline c2=cc=(ab)(ab)|\vec{c}|^2 = \vec{c} \cdot \vec{c} = (-\vec{a} - \vec{b}) \cdot (-\vec{a} - \vec{b}) \newline =aa+2ab+bb= \vec{a} \cdot \vec{a} + 2\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{b} \newline =a2+2ab+b2= |\vec{a}|^2 + 2\vec{a} \cdot \vec{b} + |\vec{b}|^2 \newline =1+2ab+1= 1 + 2\vec{a} \cdot \vec{b} + 1 \newline =2+2ab= 2 + 2\vec{a} \cdot \vec{b}
Since c2=4|\vec{c}|^2 = 4: \newline 4=2+2ab4 = 2 + 2\vec{a} \cdot \vec{b} \newline 2ab=22\vec{a} \cdot \vec{b} = 2 \newline ab=1\vec{a} \cdot \vec{b} = 1

Calculating final dot product: \newline bc=ab1=11=2\vec{b} \cdot \vec{c} = -\vec{a} \cdot \vec{b} - 1 = -1 - 1 = -2

Finding the angle: \newline bc=bccosθ\vec{b} \cdot \vec{c} = |\vec{b}||\vec{c}|\cos\theta
2=1×2×cosθ-2 = 1 \times 2 \times \cos\theta
2=2cosθ-2 = 2\cos\theta
cosθ=1\cos\theta = -1
When cosθ=1\cos\theta = -1, θ=180°\theta = 180° (or π\pi radians)

Therefore, the angle between vectors b\vec{b} and c\vec{c} is 180°180°.

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