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CBSE Class 12 Math 2026 All Sets Solved Paper

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Question : 15 of 20
Marks: +1, -0
If ∣a⃗∣=8,∣b⃗∣=3\left|\vec{a}\right|=8, \left|\vec{b}\right|=3 and ∣a⃗×b⃗∣=12\left|\vec{a}\times \vec{b}\right|=12, then the value of ∣a⃗⋅b⃗∣\left|\vec{a}\cdot \vec{b}\right|
Solution:  
∣a⃗∣=8,∣b⃗∣=3,∣a⃗×b⃗∣=12\left|\vec{a}\right|=8, \left|\vec{b}\right|=3, \left|\vec{a}\times \vec{b}\right|=12
∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ\left|\vec{a}\times \vec{b}\right| = \left|\vec{a}\right|\left|\vec{b}\right| \sin \theta
12=8×3×sin⁡θ12 = 8 \times 3 \times \sin \theta
sin⁡θ=1224=12\sin \theta = \frac{12}{24} = \frac{1}{2}
Now, cos⁡2θ=1−sin⁡2θ\cos^{2} \theta = 1 - \sin^{2} \theta
=1−14= 1 - \frac{1}{4}
=34= \frac{3}{4}
∴cos⁡θ=±32\therefore \cos \theta = \pm \frac{\sqrt{3}}{2}
a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡θ\vec{a} \cdot \vec{b} = \left|\vec{a}\right|\left|\vec{b}\right| \cos \theta
=8×3×32= 8 \times 3 \times \frac{\sqrt{3}}{2}
=123= 12 \sqrt{3}
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