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Question

Mathematics Question on Vector Algebra

If a=2i^+2j^+3k^\vec{a}=2\hat{i}+2\hat{j}+3\hat{k},b=i^+2j^+k^b=-\hat{i}+2\hat{j}+\hat{k} and c=3i^+j^c=3\hat{i}+\hat{j} are such that a+λb\vec{a}+λ\vec{b} is perpendicular to c\vec{c} ,then find the value of λλ.

Answer

The correct answer is: 8
The given vectors are a=2i^+2j^+3k^\vec{a}=2\hat{i}+2\hat{j}+3\hat{k},b=i^+2j^+k^b=-\hat{i}+2\hat{j}+\hat{k},and c=3i^+j^c=3\hat{i}+\hat{j}
Now,
a+λb\vec{a}+λ\vec{b}
=(2\hat{i}+2\hat{j}+3\hat{k})$$+\lambda(-\hat{i}+2\hat{j}+\hat{k})
=(2λ)i^+((2+2λ)j^+(3+λ)k^=(2-λ)\hat{i}+((2+2λ)\hat{j}+(3+λ)\hat{k}
if (a+λb)(\vec{a}+λ\vec{b}) is perpendicular to c\vec{c}, then
(a+λb).c=0(\vec{a}+λ\vec{b}).\vec{c}=0
[(2λ)i^+(2+2λ)j^+(3+λ)k^].(3i^+j^)=0⇒[(2-λ)\hat{i}+(2+2λ)\hat{j}+(3+λ)\hat{k}].(3\hat{i}+\hat{j})=0
(2λ)3+(2+2λ)1+(3+λ)0=0⇒(2-λ)3+(2+2λ)1+(3+λ)0=0
63λ+2+2λ=0⇒6-3λ+2+2λ=0
λ+8=0⇒-λ+8=0
λ=8⇒λ=8
Hence,the required value of λλ is 8.