?nswer it

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$$\begin{gathered} Given Vectors are \\ \stackrel{\rightharpoonup }{ a }=\hat{i}+4\hat{j}+2\hat{k}, \\ \stackrel{\rightharpoonup }{ b }=3\hat{i}-2\hat{j}+7\hat{k} and \\ \stackrel{\rightharpoonup }{ c }=2\hat{i}-\hat{j}+4\hat{k}. \\ To Calculate : A vector \stackrel{\rightharpoonup }{ d } which is perpendicular to both \stackrel{\rightharpoonup }{ a }and \stackrel{\rightharpoonup }{ b } and \stackrel{\rightharpoonup }{ c }.\stackrel{\rightharpoonup }{ d }=15 \\ solution : let the vector \stackrel{\rightharpoonup }{ d }be x\hat{i}+y\hat{j}+z\hat{k} \\ Since \stackrel{\rightharpoonup }{ d } is perpendicular to \stackrel{\rightharpoonup }{ a } and \stackrel{\rightharpoonup }{ b } \\ \Rightarrow \stackrel{\rightharpoonup }{ a }.\stackrel{\rightharpoonup }{ d }=0 and \stackrel{\rightharpoonup }{ b }.\stackrel{\rightharpoonup }{ d }=0 \\ \Rightarrow (\hat{i}+4\hat{j}+2\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=0 and (3\hat{i}-2\hat{j}+7\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=0 \\ \Rightarrow x+4y+2z=0............................\overline{)1} \\ and 3x-2y+7z=0.........................\overline{)2} \\ It is also given that \stackrel{\rightharpoonup }{ c }.\stackrel{\rightharpoonup }{ d }=15 \\ \Rightarrow (2\hat{i}-\hat{j}+4\hat{k}).(x\hat{i}+y\hat{j}+z\hat{k})=15 \\ \Rightarrow 2x-y+4z=15 \\ \Rightarrow y=2x+4z-15.............................\overline{)3} \\ Putting the value of \overline{)3} in\overline{)1} \\ \Rightarrow x+4(2x+4z-15)+2z=0 \\ \Rightarrow x+8x+16z-60+2z=0 \\ \Rightarrow 9x+18z=60 \\ dividing whole equation by 3 \\ \Rightarrow 3x+6z=20.................................\overline{)4} \\ Putting the value of \overline{)3} in\overline{)2} \\ \Rightarrow 3x-2(2x+4z-15)+7z=0 \\ \Rightarrow 3x-4x-8z+30+7z=0 \\ \Rightarrow -x-z+30 \\ \Rightarrow x+z=30 \\ \Rightarrow x=30-z...................................\overline{)5} \\ Putting the value of \overline{)5} in \overline{)4} \\ \Rightarrow 3(30-z)+6z=20 \\ \Rightarrow 90-3z+6z=20 \\ \Rightarrow 3z=20-90 \\ \Rightarrow z=\frac{-70}{3}....................................\overline{)6} \\ Putting the value of \overline{)6} in \overline{)5} \\ \Rightarrow x=30-\left(\frac{-70}{3}\right)=30+\frac{70}{3}=\frac{90+70}{3} \\ \Rightarrow x=\frac{160}{3}........................................\overline{)7} \\ Putting the value of \overline{)6} and \overline{)7} in \overline{)3} \\ \Rightarrow y=2\left(\frac{160}{3}\right)+4\left(\frac{-70}{3}\right)-15 \\ \Rightarrow y=\frac{320}{3}-\frac{280}{3}-15 \\ \Rightarrow y=\frac{320-280-45}{3}=\frac{-5}{3} \\ Hence, vector \stackrel{\rightharpoonup }{ d } which is perpendicular to both \stackrel{\rightharpoonup }{ a }and \stackrel{\rightharpoonup }{ b } and \stackrel{\rightharpoonup }{ c }.\stackrel{\rightharpoonup }{ d }=15 is \stackrel{\rightharpoonup }{ d }=\frac{160}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{70}{3}\hat{k}. \end{gathered}$$
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