You are given the two vectors a=(1,2,1)\mathbf{a}=(1,2,1)a=(1,2,1) and b=(3,−4,2)\mathbf{b}=(3,-4,2)b=(3,−4,2) . Suppose that the pair of unit vectors perpendicular to both a and b are given by
±1p(q,1,r),\pm \frac{1}{\sqrt{p}}(q,1,r),±p1(q,1,r),
where p, q and r are some constants.
Determine the values of p, q and r.
c=a×b=∣ijk1213−42∣=i(2⋅2−(−4)⋅1)−j(1⋅2−1⋅3)+k(1⋅(−4)−3⋅2)=8i+j−10k=1p(q,1,r)8=qp1=1p∴p=1q=81=8−10=rpr=−101=−10c=\bold{a}\times\bold{b}=\begin{vmatrix} i & j & k \\ 1 & 2 & 1\\ 3 & -4 & 2 \end{vmatrix}\\ =i(2\cdot2-(-4)\cdot1)-j(1\cdot2-1\cdot3)+k(1\cdot(-4)-3\cdot2)=8i+j-10k\\ =\dfrac{1}{\sqrt{p}}(q,1,r)\\ 8=\dfrac{q}{\sqrt{p}}\\ 1=\dfrac{1}{\sqrt{p}}\\ \therefore p=1\\ q=8\sqrt{1}=8\\ -10=\dfrac{r}{\sqrt{p}}\\ r=-10\sqrt{1}=-10c=a×b=∣∣i13j2−4k12∣∣=i(2⋅2−(−4)⋅1)−j(1⋅2−1⋅3)+k(1⋅(−4)−3⋅2)=8i+j−10k=p1(q,1,r)8=pq1=p1∴p=1q=81=8−10=prr=−101=−10
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