2. (a) Prove that āx(5 x c) = (a.c)...
Question

# 2. (a) Prove that āx(5 x c) = (a.c)ā - (ā.bjë.

11th - 12th Class
Maths
Solution
102
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Proof. ( (i) ) Let ( vec{a}=a_{1} hat{i}+a_{2} hat{j}+a_{3} ) [ begin{array}{l} vec{b}=b_{1} hat{i}+b_{2} hat{j}+b_{3} hat{k} vec{c}=c_{1} hat{i}+c_{2} hat{j}+c_{3} hat{k} end{array} ] ( therefore quad vec{b} times vec{c}=left|begin{array}{ccc}hat{i} & hat{j} & hat{k} b_{1} & b_{2} & b_{3} c_{1} & c_{2} & c_{3}end{array}right| ) ( =left(b_{2} c_{3}-b_{3} c_{2}right) hat{i}+left(b_{3} c_{1}-b_{1} c_{3}right) hat{j}+left(b_{1} c_{2}-b_{2} c_{1}right) hat{k} ) Now, ( vec{a} times(vec{b} times vec{c})=left|begin{array}{ccc}hat{i} & hat{j} & hat{k} a_{1} & a_{2} & a_{3} b_{2} c_{3}-b_{3} c_{2} & b_{3} c_{1}-b_{1} c_{3} & b_{1} c_{2}-b_{2} c_{1}end{array}right| ) ( =left(a_{2} b_{1} c_{2}-a_{2} b_{2} c_{1}-a_{3} b_{3} c_{1}+a_{3} b_{1} c_{3}right) hat{i}+left(a_{3} b_{2} c_{3}-a_{3} b_{3} c_{2}-a_{1} b_{1} c_{2}+c_{3}right. ) ( +left(a_{1} b_{3} c_{1}-a_{1} b_{1} c_{3}-a_{2} b_{2} c_{3}+a_{2} b_{3} c_{2}right) hat{k}_{k} ) [Expanding along first Also [ begin{array}{l} vec{a} cdot vec{b}=a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3} vec{a} cdot vec{c}=a_{1} c_{1}+a_{2} c_{2}+a_{3} c_{3} end{array} ] Now, ( (vec{a} cdot vec{c}) vec{b}-(vec{a} cdot vec{b}) vec{c}=left(a_{1} c_{1}+a_{2} c_{2}+a_{3} c_{3}right)left(b_{1} hat{i}+b_{2} hat{j}+b_{3} hat{k}right) ) [ begin{array}{c} -left(a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}right)left(c_{1} hat{i}+c_{2} hat{j}+c_{3} hat{k}right) =left(a_{1} b_{1} c_{1}+a_{2} b_{1} c_{2}+a_{3} b_{1} c_{3}-a_{1} b_{1} c_{1}-a_{2} b_{2} c_{1}-a_{3} b_{3} c_{1}right) hat{i} +left(a_{1} b_{2} c_{1}+a_{2} b_{2} c_{2}+a_{3} b_{2} c_{3}-a_{1} b_{1} c_{2}-a_{2} b_{2} c_{2}-a_{3} b_{3} c_{2}right) hat{j} +left(a_{1} b_{3} c_{1}+a_{2} b_{3} c_{2}+a_{3} b_{3} c_{3}-a_{1} b_{1} c_{3}-a_{2} b_{2} c_{3}-a_{3} b_{3} c_{3}right) =left(a_{2} b_{1} c_{2}+a_{3} b_{1} c_{3}-a_{2} b_{2} c_{1}-a_{3} b_{3} c_{1}right) hat{i} +left(a_{1} b_{2} c_{1}+a_{3} b_{2} c_{3}-a_{1} b_{1} c_{2}-a_{3} b_{3} c_{2}right) hat{j} +left(a_{1} b_{3} c_{1}+a_{2} b_{3} c_{2}-a_{1} b_{1} c_{3}-a_{2} b_{2} c_{3}right) hat{k} end{array} ]