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If ∣a⃗∣=10,∣b⃗∣=2|\vec{a}|=10,|\vec{b}|=2 and a⃗⋅b⃗=12\vec{a} \cdot \vec{b}=12, then the value of ...
Feb 3, 2024
If ∣a⃗∣=10,∣b⃗∣=2|\vec{a}|=10,|\vec{b}|=2 and a⃗⋅b⃗=12\vec{a} \cdot \vec{b}=12, then the value of ∣a⃗×b⃗∣|\vec{a} \times \vec{b}| is (a) 5 (b) 10 (c) 14 (d) 16
Solution by Steps
step 2
The magnitude of the cross product of two vectors can be found using the formula ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡(θ)|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\theta), where θ\theta is the angle between a⃗\vec{a} and b⃗\vec{b}
step 3
The dot product of two vectors is given by a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡(θ)\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos(\theta). We can solve for cos⁡(θ)\cos(\theta) using the given dot product: cos⁡(θ)=a⃗⋅b⃗∣a⃗∣∣b⃗∣=1210×2=1220=35\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{12}{10 \times 2} = \frac{12}{20} = \frac{3}{5}
step 4
To find sin⁡(θ)\sin(\theta), we use the identity sin⁡2(θ)+cos⁡2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1. Thus, sin⁡(θ)=1−cos⁡2(θ)=1−(35)2=1−925=1625=45\sin(\theta) = \sqrt{1 - \cos^2(\theta)} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}
step 5
Now we can find the magnitude of the cross product: ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡(θ)=10×2×45=20×45=16|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\theta) = 10 \times 2 \times \frac{4}{5} = 20 \times \frac{4}{5} = 16
D
Key Concept
Magnitude of Cross Product
Explanation
The magnitude of the cross product of two vectors is equal to the product of their magnitudes and the sine of the angle between them.
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