If F⃗=2î+ĵ-k̂ and r⃗=3î+2ĵ-2k̂, then the scalar and vector products of F⃗ and r⃗ have the magnitudes respectively as

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(4) 10, √2. Given: F⃗=2î+ĵ-k̂ and r⃗=3î+2ĵ-2k̂. Idea: take the dot product term by term, then the cross product as a determinant. F⃗·r⃗=(2)(3)+(1)(2)+(-1)(-2)=6+2+2=10. F⃗×r⃗=î[(1)(-2)-(-1)(2)]-ĵ[(2)(-2)-(-1)(3)]+k̂[(2)(2)-(1)(3)]. =î(-2+2)-ĵ(-4+3)+k̂(4-3)=ĵ+k̂. |F⃗×r⃗|=√1²+1²=√2. So the magnitudes are 10 and √2.

If F⃗=2î+ĵ-k̂ and r⃗=3î+2ĵ-2k̂, then the scalar and vector products of F⃗ and r⃗ have the magnitudes respectively as

Step-by-step solution

Why the other options are wrong