Let S be the set of all ( , ) for which the vectors i-j+k,i+2j+ k and 3i-4j+5k, where - =5…

Question

Let $S$ be the set of all $(\lambda ,\mu )$ for which the vectors $\lambda \hat{i}-\hat{j}+\hat{k},\hat{i}+2\hat{j}+\mu \hat{k}$ and $3\hat{i}-4\hat{j}+5\hat{k}$, where $\lambda -\mu =5$, are coplanar, then $\sum _{(\lambda ,\mu )\in S}80({\lambda }^{2}+{\mu }^{2})$ is equal to :

Accepted Answer

Given condition for coplanarity

For three vectors to be coplanar, their scalar triple product must be zero. We have the vectors A, B, and C, and we know the given relation between \lambda  and \mu :

$A=\lambda \hat{i}-\hat{j}+\hat{k}$

$B=\hat{i}+2\hat{j}+\mu \hat{k}$

$C=3\hat{i}-4\hat{j}+5\hat{k}$

Scalar triple product condition:

$[A,B,C]=A\cdot (B	imes C)=0$

Calculate the cross product of B and C

$B	imes C=\begin{vmatrix} \hat{i} \quad  \hat{j} \quad  \hat{k} \ 1 \quad  2 \quad  \mu \ 3 \quad  -4 \quad  5 \end{vmatrix}$

$B	imes C=(10+4\mu )\hat{i}-(5-3\mu )\hat{j}-10\hat{k}$

Calculate the scalar triple product and apply the coplanarity condition

$[A,B,C]=A\cdot (B	imes C)=\lambda (10+4\mu )-1(5-3\mu )+1(-10)=0$

Substitute the given relation between \lambda  and \mu 
Given that:

$\lambda -\mu =5$

From the coplanarity condition, we have:

$4\lambda \mu +10\lambda -3\mu =5$

Now, substitute \lambda  in terms of \mu :

$4(5+\mu )\mu +10(5+\mu )-3\mu =5$

Solve for \mu  and \lambda

Simplify the equation and solve for \mu :

$4{\mu }^{2}+27\mu +45=0$

Factor the equation:

$(4\mu +15)(\mu +3)=0$

So, the two possible values for \mu  are:

$\mu =-3,\frac{-15}{4}$

For each value of \mu , find the corresponding value of \lambda  using the given relation:

$\lambda =\mu +5$

So, we get the values for \lambda :

$\lambda =2,\frac{5}{4}$

Calculate the sum

Now, we need to find the sum:

$\sum _{(\lambda ,\mu )\in S}80({\lambda }^{2}+{\mu }^{2})$

Substitute the values of \lambda  and \mu , and simplify:

$80[({2}^{2}+(-3{)}^{2})+{(\frac{5}{4})}^{2}+{(\frac{-15}{4})}^{2}]=80[13+\frac{25}{16}+\frac{225}{16}]$

$=80[13+\frac{250}{16}]=10	imes 229$

$=2290$

So, the correct answer is 2290 (Option D).

Solutions · 1