Determine the surface area of a parallelogram constructed on 2 vectors

Question

Please, help me with this.

Determine the surface area of a parallelogram constructed on 2 vectors given as adjacent sides: $ v1 = a + 2b $ and $ v2 = a - 3b $ ( 'a' and 'b' being also vectors )

Also: $ |a| = 5 $ , $|b| = 3 $ , $m(a,b) = pi/6$

Now, I know the formula for the surface area or a parallelogram constructed on vectors is: $ |v1| * |v2| * sin( m(v1,v2) ) $.

Normally, this would be an easy exercise for me, if the vectors v1, and v2 would be in the general $xi + yj + zk = 0$ form, but they are constructed upon other vectors: a and b.

I tried defining the a and b vectors as such: $ a = x1i + y1j + z1k $ $ b = x2i + y2j + z2k $ but I couldn't finish the calculus.

How can I solve this? Please someone guide me since most of my future exam exercises involve such vectors.

Answer 1

The area of the parallelogram is the (absolute value of the) cross product of the vectors that make up the edges.

The cross product distributes over addition.

$v_1\times v_2 = (a+2b)\times(a+ (-3b)) = a\times a + a\times (-3b) + 2b\times a + 2b\times (-3b)$

What else do you need to know?

the cross product is anti-commutative. $a\times b = -b\times a$

Which implies that $a\times a = 0$

And we can factor out scalar multiples. $(2a)\times b = 2(a\times b)$

$|v_1\times v_2| = |5(a\times b)|$

Answer 2

HINT

Use the cross product, since

$$|\vec v_1 \times \vec v_2|=|\vec v_1||\vec v_2|\sin \theta$$

and

$$v_1=(x_1,y_1,z_1)\;,\;\;\;v_2=(x_2,y_2,z_2)$$

$$v_1\times v_2:=\begin{vmatrix}\vec i&\vec j&\vec k\\x_1&y_1&z_1\\x_2&y_2&z_2\end{vmatrix}=(y_1z_2-z_1y_2\,,\,x_2z_1-x_1z_2\,,\,x_1y_2-x_2y_1)$$