Vector (Cross) Product - Celestial Tutors

Vector(Cross) Product

Let $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ be two non zero, non parallel vectors. Then the vector product $\dpi{120} \overrightarrow{a}X\overrightarrow{b}$ is defined as,

$\dpi{150} \mathbf{\overrightarrow{a}x\overrightarrow{b}=\left | a \right |\left | b \right |sin\theta}$

Where θ is the angle between vectors a and b.

Properties of vector product:

If cross product of two vectors is zero that means vectors are parallel.

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=0 \Rightarrow$ parallel vectors

Vector product is not commutative.

$\dpi{120} \overrightarrow{a}X\overrightarrow{b} =-\left ( \overrightarrow{b}X\overrightarrow{a} \right )$

Vector Product is distributive

$\dpi{120} \overrightarrow{a}X\left ( \overrightarrow{b}+\overrightarrow{c} \right )=\overrightarrow{a}X\overrightarrow{b}+\overrightarrow{a}X\overrightarrow{c}$

$\dpi{120} \left ( \overrightarrow{b}+\overrightarrow{c} \right )X\overrightarrow{a}= \overrightarrow{b}X\overrightarrow{a}+\overrightarrow{c}X\overrightarrow{a}$

Some important applications of vector product

Area of parallelogram with adjacent sides $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ is

$\dpi{120} \mathbf{\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

Area of a triangle with adjacent sides $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ is

$\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

Area of parallelogram with diagonals $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ is

$\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

Volume of parallelepiped is given as

$\dpi{120} \mathbf{\left | \overrightarrow{a}.\left ( \overrightarrow{b}X\overrightarrow{c} \right ) \right |}$

Three vectors are coplanar if their scalar triple product is 0 i.e $\dpi{120} \mathbf{ \overrightarrow{a}.\left ( \overrightarrow{b}X\overrightarrow{c} \right ) }=0$ or three vector are coplanar if volume of parallelepiped formed by them is 0.

Example1. Find $\dpi{120} {\color{Red} \left | \overrightarrow{a}X\overrightarrow{b} \right |}$ for given vectors $\dpi{120} {\color{Red} \overrightarrow{a}}$ =i+3j-2k and $\dpi{120} {\color{Red} \overrightarrow{b}}$ =-i+3k.

Solution:

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=\begin{vmatrix} i & j & k \\ 1& 3 &-2 \\ -1& 0 & 3 \end{vmatrix}$

= i(9-0) –j(3-2) +k(0+3)

= 9i-j+3k

$\dpi{120} \left | \overrightarrow{a}X\overrightarrow{b} \right |= \sqrt{(9)^{2}+(-1)^{2}+(3)^{2}} =\sqrt{91}$

Example2. A plane is defined by any three points that are in the plane. If a plane contains the points P=(1,0,0), Q=(1,1,1) and R=(2,−1,3) find a vector that is orthogonal to the plane.

Solution : To find orthogonal(normal) vector of a plane, we just find the cross product of two vectors lying in that plane.

So first we find two vectors using three points lying in the plane.

$\dpi{120} \overrightarrow{PQ}$ = (1,1,1) –(1,0,0)= 〈0,1,1〉

$\dpi{120} \overrightarrow{PR}$ =(2,−1,3)–(1,0,0)= 〈1,-1,3〉

Orthogonal vector is,

$\dpi{120} \overrightarrow{PQ}X\overrightarrow{PR}=\begin{vmatrix} i & j & k \\ 0& 1 &1 \\ 1& -1 & 3 \end{vmatrix}$

= i(3+1) –j(0-1) +k(0-1) = 4i +j –k

Example3. Find a unit vector of magnitude 9, perpendicular to both the vectors = i-2j+3k and = i+2j-k.

Solution:

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=\begin{vmatrix} i & j & k \\ 1& -2 &3 \\ 1& 2 & -1 \end{vmatrix}$

= i(2-6) –j(-1-3) +k(2+2)

= i(-4)-j(-4) +4k

= -4i+4j+4k

$\dpi{120} \left | \overrightarrow{a}X\overrightarrow{b} \right |= \sqrt{(-4)^{2}+(4)^{2}+(4)^{2}} =\sqrt{48}=4\sqrt{3}$

Required unit vector

$\dpi{120} =9\frac{\overrightarrow{a}X\overrightarrow{b}}{\left | \overrightarrow{a}X\overrightarrow{b} \right |}$ (multiply with given magnitude 9)

$\dpi{120} =9\frac{\left \langle -4,4,4 \right \rangle}{4\sqrt{3}}=\frac{3\sqrt{3}\left \langle -4,4,4 \right \rangle}{4}= 3\sqrt{3}\left \langle -1,1,1 \right \rangle$

Example4. Show that area of parallelogram having diagonals 〈3,1,-2〉 and 〈1,-3,4〉 is 5√3 .

Solution: lets assume these diagonals are represented by vectors $\dpi{120} \overrightarrow{a}$ and $\dpi{120} \overrightarrow{b}$ .

$\dpi{120} \overrightarrow{a}X\overrightarrow{b}=\begin{vmatrix} i & j & k \\ 3& 1 &-2 \\ 1& -3 & 4 \end{vmatrix}$

= i(4-6) -j(12+2) +k(-9-1)

= -2i-14j-10k

$\dpi{120} \left | \overrightarrow{a}X\overrightarrow{b} \right |= \sqrt{(-2)^{2}+(-14)^{2}+(-10)^{2}} =\sqrt{300}$

Area of parallelogram = $\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{a}X\overrightarrow{b} \right |}$

= 1/2 (√300) = 5√3

Example5.Find area of triangle whose vertices are A(3,-1,2), B(1,-1,-3) and C(4,-3,1).

Solution: First we find vectors using given points.

$\dpi{120} \overrightarrow{AB}$ = (1,-1,-3) –(3,-1,2)= 〈-2,0,-5〉

$\dpi{120} \overrightarrow{AC}$ = (4,−3,1)–(3,-1,2)= 〈1,-2,-1〉

$\dpi{120} \overrightarrow{AB}X\overrightarrow{AC}=\begin{vmatrix} i & j & k \\ -2& 0 &-5 \\ 1& -2 & -1 \end{vmatrix}$

=i(0-10)-j(2+5)+k(4-0)

= -10i-7j+4k

$\dpi{120} \left | \overrightarrow{AB}X\overrightarrow{AC} \right |= \sqrt{(-10)^{2}+(-7)^{2}+(4)^{2}} =\sqrt{165}$

Area of triangle ABC

$\dpi{120} \mathbf{\frac{1}{2}\left | \overrightarrow{AB}X\overrightarrow{AC} \right |}= \frac{1}{2}\sqrt{165}$

Example6. Determine if three vectors lie in same plane (coplanar) or not.

a=〈1,4,-7〉 , b=〈2,-1,4〉 , c =〈0,-9,18〉

Solution:

$\dpi{120} \overrightarrow{a}.\left ( \overrightarrow{b}X\overrightarrow{c} \right )= \begin{vmatrix} 1 &4 &-7 \\ 2& -1 & 4\\ 0& -9 & 18 \end{vmatrix}$

= 1(-18+36)-4(36-0)-7(-18-0)

= 18 – 144 +126 = 0

Since scalar triple product is 0 i.e volume of parallelepiped is 0 so all three vectors lie in same plane.

Practice problems:

Find a unit vector of magnitude 3, perpendicular to both the vectors a= 3i+j-4k and b= 6i+5j-2k.
Find the area of parallelogram determined by vectors 〈3,1,-2〉 and 〈1,-3,4〉
Determine of three vectors lie in same plane (coplanar) or not.

a=〈2,3,1〉 , b=〈1,-1,0〉 , c =〈7,3,2〉

Let a=i+4j+2k , b=3i-2j+7k and c=2i-j+4k. Find a vector d which is perpendicular to both a and b and c.d =15

Answers:

2i-2j+k
(5√3)/2 units
Yes, coplanar
1/3 (2i-2j-k)