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Question Number 132856 by bramlexs22 last updated on 17/Feb/21
Given vector a^→ = i^� +j^� +k^� , c^→ =j^� −k^� ; a^→ × b^→ = c^→ and a^→ .b^→ = 3 then ∣b^→ ∣ = ?
$$\mathrm{Given}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:=\:\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}}\:,\:\overset{\rightarrow} {\mathrm{c}}=\hat {\mathrm{j}}−\hat {\mathrm{k}}\:; \\ $$$$\:\overset{\rightarrow} {\mathrm{a}}\:×\:\overset{\rightarrow} {\mathrm{b}}\:=\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{and}\:\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\:=\:\mathrm{3}\:\mathrm{then}\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:? \\ $$
Answered by EDWIN88 last updated on 17/Feb/21
from a^→ × b^→ = c^→ we get a^→ ×(a^→ ×b^→ ) = a^→ ×c^→ a^→ ×c^→ = determinant (((1 1 1)),((0 1 −1)))= −2i^� +j^� +k^� and a^→ ×(a^→ ×b^→ )=(a^→ .b^→ )a^→ −(a^→ .a^→ )b^→ =−2i^� +j^� +k^� ⇒3a^→ −3b^→ = −2i^� +j^� +k^� b^→ =(((3,3,3)−(−2,1,1))/3) = ((5i^� +2j^� +2k^� )/3) Therefore ∣b^→ ∣ = ((√(25+4+4))/3) = ((√(33))/3) = (√((11)/3))
$$\mathrm{from}\:\overset{\rightarrow} {\mathrm{a}}\:×\:\overset{\rightarrow} {\mathrm{b}}\:=\:\overset{\rightarrow} {\mathrm{c}}\:\mathrm{we}\:\mathrm{get}\:\overset{\rightarrow} {\mathrm{a}}×\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)\:=\:\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{c}} \\ $$$$\:\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{c}}=\begin{vmatrix}{\mathrm{1}\:\:\:\mathrm{1}\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{0}\:\:\:\mathrm{1}\:\:−\mathrm{1}}\end{vmatrix}=\:−\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {\mathrm{a}}×\left(\overset{\rightarrow} {\mathrm{a}}×\overset{\rightarrow} {\mathrm{b}}\right)=\left(\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{b}}\right)\overset{\rightarrow} {\mathrm{a}}−\left(\overset{\rightarrow} {\mathrm{a}}.\overset{\rightarrow} {\mathrm{a}}\right)\overset{\rightarrow} {\mathrm{b}}=−\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}} \\ $$$$\Rightarrow\mathrm{3}\overset{\rightarrow} {\mathrm{a}}−\mathrm{3}\overset{\rightarrow} {\mathrm{b}}=\:−\mathrm{2}\hat {\mathrm{i}}+\hat {\mathrm{j}}+\hat {\mathrm{k}}\: \\ $$$$\:\overset{\rightarrow} {\mathrm{b}}=\frac{\left(\mathrm{3},\mathrm{3},\mathrm{3}\right)−\left(−\mathrm{2},\mathrm{1},\mathrm{1}\right)}{\mathrm{3}}\:=\:\frac{\mathrm{5}\hat {\mathrm{i}}+\mathrm{2}\hat {\mathrm{j}}+\mathrm{2}\hat {\mathrm{k}}}{\mathrm{3}} \\ $$$$\mathrm{Therefore}\:\mid\overset{\rightarrow} {\mathrm{b}}\mid\:=\:\frac{\sqrt{\mathrm{25}+\mathrm{4}+\mathrm{4}}}{\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{33}}}{\mathrm{3}}\:=\:\sqrt{\frac{\mathrm{11}}{\mathrm{3}}} \\ $$

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