Question

If $a=2i+j-3k,b=i-2j+k$ $c=-i+j-4k$ and $d=i+j+k$, then $|(a\times b)\times (c\times d)|=$

• A. $5\sqrt{114}$
• B. $5\sqrt{94}$
• C. $5\sqrt{124}$
• D. $5\sqrt{104}$

Answer 1

Answer: (A)

Given that,$a=2\hat{i}+\hat{j}-3\hat{k}$
$b=\hat{i}-2\hat{j}+\hat{k}$
$c=-\hat{i}+\hat{j}-4\hat{k}$ and $d=\hat{i}+\hat{j}+\hat{k}$
Now, $a\times b=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 1& -3\\ 1& -2& 1\end{array}\right|$
$=i(1-6)-\hat{j}(2+3)+\hat{k}(-4-1)$
$a\times b=-5\hat{i}-5\hat{j}-5\hat{k}$
Now, $c\times d=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -1& 1& -4\\ 1& 1& 1\end{array}\right|$
$=\hat{i}(1+4)-\hat{j}(-1+4)+\hat{k}(-1-1)$
$=5\hat{i}-3\hat{j}-2\hat{k}$
Now, $(a\times b)\times (c\times d)=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ -5& -5& -5\\ 5& -3& -2\end{array}\right|$
$=\hat{i}(10-15)-\hat{j}(10+25)+\hat{k}(+15+25)$
$=-5\hat{i}-35\hat{j}+40\hat{k}=5(-\hat{i}-7\hat{j}+8\hat{k})$
$|(a\times b)\times (c\times d)|=5\sqrt{(-1{)}^2+(-7{)}^2+(8{)}^2}$
$=5\sqrt{1+49+64}=5\sqrt{114}$