Question

If a = i - 2j + k, b = 2i + j + k, c = i + 2j - k, find a × (b × c) and |(a × b) × c|.

Answer 1

Answer:

| a×(b×c)|=  $\sqrt{174}$

a × (b × c)$=-9\hat i -6\hat j-3\hat k$

Step-by-step explanation:

$\vec a=\hat i-2\hat j+\hat k\\\\\vec b=2\hat i+\hat j+\hat k\\\\\\\vec a \times \vec b=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&-2&1\\2&1&1\end{array}\right| \\\\=\hat i(-2-1)-\hat j(1-2)+\hat k(1+4)\\\\=-3\hat i+\hat j+5\hat k$

Now for  

$(\vec a \times\vec b)\times \vec c=(-3\hat i+\hat j+5\hat k)\times(\hat i+2\hat j-\hat k)\\\\( \vec a\times\vec b)\times \vec c=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\-3&1&5\\1&2&-1\end{array}\right|\\\\\\=\hat i(-1-10)-\hat j(3-5)+\hat k(-6-1)\\\\\\=-11\hat i +2\hat j-7\hat k$

For magnitude of  

| a×(b×c)|=

$\sqrt{(-11)^{2}+2^{2}+(-7)^{2}   } \\\\=\sqrt{121+4+49} \\\\=\sqrt{174}$

For calculation of a × (b × c): first calculate  (b × c)

$=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\2&1&1\\1&2&-1\end{array}\right|\\\\\\=\hat i(-1-2)-\hat j(-2-1)+\hat k(4-1)\\\\\\=-3\hat i +3\hat j+3\hat k$

a × (b × c)

$=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\1&-2&1\\-3&3&3\end{array}\right|\\\\\\=\hat i(-6-3)-\hat j(3+3)+\hat k(3-6)\\\\\\=-9\hat i -6\hat j-3\hat k$

Answer 2

Step-by-step explanation:

solutions -9i -6j -3k. ; √174