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Question Number 15352 by Tinkutara last updated on 09/Jun/17

Let a^→ = i^∧ + 4j^∧ + 2k^∧ , b^→ = 3i^∧ − 2j^∧ + 7k^∧ and c^→ = 2i^∧ − j^∧ + 4k^∧ . Find a vector d^→ which is perpendicular to both a^→ and b^→ , and c^→ ∙ d^→ = 15.

$$\mathrm{Let}\:\overset{\rightarrow} {{a}}\:=\:\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}\:+\:\mathrm{2}\overset{\wedge} {{k}},\:\overset{\rightarrow} {{b}}\:=\:\mathrm{3}\overset{\wedge} {{i}}\:−\:\mathrm{2}\overset{\wedge} {{j}}\:+\:\mathrm{7}\overset{\wedge} {{k}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{c}}\:=\:\mathrm{2}\overset{\wedge} {{i}}\:−\:\overset{\wedge} {{j}}\:+\:\mathrm{4}\overset{\wedge} {{k}}\:.\:\mathrm{Find}\:\mathrm{a}\:\mathrm{vector}\:\overset{\rightarrow} {{d}} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{both}\:\overset{\rightarrow} {{a}}\:\mathrm{and}\:\overset{\rightarrow} {{b}}, \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{c}}\:\centerdot\:\overset{\rightarrow} {{d}}\:=\:\mathrm{15}. \\ $$

Commented by prakash jain last updated on 10/Jun/17

a^→ ×b^→ = determinant ((i,j,k),(1,4,2),(3,(−2),7)) =i(28+4)−j(7−6)+k(−2−12) =32i−j−14k a^→ ×b^→ is perpendicular to both a and b. unit vector perpendicular to a and b n=((a×b)/(∣a×b∣))=((32i−j−14k)/(√(32^2 +1+14^2 )))=((32i−j−14k)/(√(1221))) d=kn=k((32i−j−14k)/(√(1221))) c∙d=15 k((32×2+1−14×4)/(√(1221)))=15 k(9/(√(1221)))=15 k=((15(√(1221)))/9) d=kn=((15(√(1221)))/9)(((32i−j−14k)/(√(1221)))) d=(5/3)(32i−j−14k)

$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}=\begin{vmatrix}{\boldsymbol{{i}}}&{\boldsymbol{{j}}}&{\boldsymbol{{k}}}\\{\mathrm{1}}&{\mathrm{4}}&{\mathrm{2}}\\{\mathrm{3}}&{−\mathrm{2}}&{\mathrm{7}}\end{vmatrix} \\ $$$$=\boldsymbol{{i}}\left(\mathrm{28}+\mathrm{4}\right)−\boldsymbol{{j}}\left(\mathrm{7}−\mathrm{6}\right)+\boldsymbol{{k}}\left(−\mathrm{2}−\mathrm{12}\right) \\ $$$$=\mathrm{32}\boldsymbol{{i}}−\boldsymbol{{j}}−\mathrm{14}\boldsymbol{{k}} \\ $$$$\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\:\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:\mathrm{both}\:\boldsymbol{{a}}\:\mathrm{and}\:\boldsymbol{{b}}. \\ $$$${unit}\:{vector}\:{perpendicular}\:{to}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}} \\ $$$$\boldsymbol{{n}}=\frac{\boldsymbol{{a}}×\boldsymbol{{b}}}{\mid\boldsymbol{{a}}×\boldsymbol{{b}}\mid}=\frac{\mathrm{32}{i}−{j}−\mathrm{14}{k}}{\sqrt{\mathrm{32}^{\mathrm{2}} +\mathrm{1}+\mathrm{14}^{\mathrm{2}} }}=\frac{\mathrm{32}{i}−{j}−\mathrm{14}{k}}{\sqrt{\mathrm{1221}}} \\ $$$$\boldsymbol{{d}}={k}\boldsymbol{{n}}={k}\frac{\mathrm{32}{i}−{j}−\mathrm{14}{k}}{\sqrt{\mathrm{1221}}} \\ $$$$\boldsymbol{{c}}\centerdot\boldsymbol{{d}}=\mathrm{15} \\ $$$${k}\frac{\mathrm{32}×\mathrm{2}+\mathrm{1}−\mathrm{14}×\mathrm{4}}{\sqrt{\mathrm{1221}}}=\mathrm{15} \\ $$$${k}\frac{\mathrm{9}}{\sqrt{\mathrm{1221}}}=\mathrm{15} \\ $$$${k}=\frac{\mathrm{15}\sqrt{\mathrm{1221}}}{\mathrm{9}} \\ $$$$\boldsymbol{{d}}={k}\boldsymbol{{n}}=\frac{\mathrm{15}\sqrt{\mathrm{1221}}}{\mathrm{9}}\left(\frac{\mathrm{32}\boldsymbol{{i}}−\boldsymbol{{j}}−\mathrm{14}\boldsymbol{{k}}}{\sqrt{\mathrm{1221}}}\right) \\ $$$$\boldsymbol{{d}}=\frac{\mathrm{5}}{\mathrm{3}}\left(\mathrm{32}\boldsymbol{{i}}−\boldsymbol{{j}}−\mathrm{14}\boldsymbol{{k}}\right) \\ $$

Commented by Tinkutara last updated on 10/Jun/17

But answer is (1/3)(160i^∧ − 5j^∧ + 70k^∧ )

$$\mathrm{But}\:\mathrm{answer}\:\mathrm{is}\:\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{160}\overset{\wedge} {{i}}\:−\:\mathrm{5}\overset{\wedge} {{j}}\:+\:\mathrm{70}\overset{\wedge} {{k}}\right) \\ $$

Commented by prakash jain last updated on 10/Jun/17

corrected. Forgot to cancel (√(1221))

$$\mathrm{corrected}.\:\mathrm{Forgot}\:\mathrm{to}\:\mathrm{cancel}\:\sqrt{\mathrm{1221}} \\ $$

Commented by Tinkutara last updated on 10/Jun/17

Thanks Sir!

$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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