Question

suppose A=3i-j-2k and B=2i+3j+k,Find |AxB|,(A+2B)x(2A-B),(A+B)x(A-B)

Answer 1

Lets write the cross product formula:

AxB =$\begin{vmatrix} i & j & k\\ A_1 & A_2 & A_3\\ B_1 & B_2 & B_3 \end{vmatrix} =$

= (A₂B₃ - A₃B₂)i - (A₁B₃ - A₃B₁)j + (A₁B₂ - A₂B₁)k

The length of vector is:

$|a|=\sqrt{a_1^2+a_2^2+a_3^2}$

The sum of two vectors is:

a + b = {a₁ + b₁; a₂ + b₂; a₃ + b₃}

The difference of two vectors is:

a - b = {a₁ - b₁; a₂ - b₂; a₃ - b₃}

So let's find |AxB|:

AxB = $\begin{vmatrix} i & j & k\\ 3 & -1 & -2\\ 2 & 3 & 1 \end{vmatrix} =$

= ((-1)*1 - (-2)*3)i - (3*1 - (-2)*2) + (3·3 - (-1)·2)k = (-1 + 6)i - (3 + 4)j + (9 + 2)k = {5; -7; 11}

|AxB| = $\sqrt{5^2+(-7)^2+11^2}=\sqrt{25+49+121}=\sqrt{195}\approx$ 13.96

Now let's find (A+2B)x(2A-B):

2B = {2*B₁; 2*B₂; 2*B₃} = {2*2; 2*3; 2*1} = {4; 6; 2}

2A = {2*A₁; 2*A₂; 2*A₃} = {2*3; 2*(-1); 2*(-2)} = {6; -2; -4}

A + 2B = {3 + 4; (-1) + 6; (-2)+2} = {7; 5; 0}

2A - B = {6 - 2; (-2) - 3; (-4) - 1} = {4; -5; -5}

(A + 2B)x(2A - B) = $\begin{vmatrix} i & j & k\\ 7 & 5 & 0\\ 4 & -5 & -5 \end{vmatrix} =$

= (5*(-5) - 0*(-5))i - (7*(-5) - 0*4)j + (7*(-5) - 5*4)k = (-25 - 0)i - (-35 - 0)j + (-35 - 20)k = {-25; 35; -55}

Finally let's find (A + B)x(A - B):

A + B = {3 + 2; (-1) + 3; (-2) + 1} = {5; 2; -1}

A - B = {3 - 2; (-1) - 3; (-2) - 1} = {1; -4; -3}

(A + B)x(A - B) = $\begin{vmatrix} i & j & k\\ 5 & 2 & -1\\ 1 & -4 & -3 \end{vmatrix} =$

= (2*(-3) - (-1)*(-4))i - (5*(-3) - (-1)*1)j + (5*(-4) - 2*1)k = (-6 - 4)i - (-15 + 1)j + (-20 - 2)k = {-10; 14; -22}