Cross product also known as the vector product is the product of the magnitude of the vectors u and v. It is denoted as u $\times$ v. The resultant cross product is perpendicular to both u and v.
Consider two vector

$\vec{u}$ = a
1i + b1 j + c1 k
$\vec{v}$ = a2 i + b2 j + c2 k

The cross product is given by
Solving the determinant you get the cross product.

Cross product Calculator is an online tool used to find the cross product of any two given vectors. Provided are blocks where you are supposed to enter the magnitude of the components of the given vector to get the cross product.

## Cross Product Problems

Here are some steps for Cross product
Step 1 : Read the problem and note down the given vectors of the form
$\vec{u}$ = a1 i + b1 j + c1 k
$\vec{v}$ = a2 i + b2 j + c2 k.

Step 2 : Write the matrix form Using given vectors and then calculate the determinant of it. Solve the determinant as below to get the answer
u $\times$ v = $\begin{vmatrix} a_{1} & b_{1} & c_{1}\\ a_{2} & b_{2} & c_{2}\\ a_{3} & b_{3} & c_{3} \end{vmatrix}$
= a1 (b2c3 - b3 c2) - b1 (a2c3 - a3c2) + c1 (a2b3 - a3b2)

Substituting the values in the determinant to get the answer.

Below are given some solved problems on cross product which may be helpful for you.

### Solved Examples

Question 1: Find the cross product of u = (3, 5, 7) and v = (-1, 4, 2).
Solution:

Step 1: The given vectors are
$\vec{u}$ = 3i + 5j + 7k
$\vec{v}$ = -i + 4j + 2k

Step 2: The cross product is given by
u $\times$ v = $\begin{vmatrix} i & j & k\\ 3 & 5 & 7\\ -1 & 4 & 2 \end{vmatrix}$
= i (5(2) - 4(7)) - j (3(2) - (-1)7) + k (3(4) - (-1)5)
= -18i - 13j + 17k.

Question 2: Find the cross product of u = (0, 5, 1) and v = (-1, 0, 2).
Solution:

Step 1: The given vectors are
$\vec{u}$ = 5j + k
$\vec{v}$ = -i + 2k

Step 2: The cross product is given by
u $\times$ v = $\begin{vmatrix} i & j & k\\ 0 & 5 & 1\\ -1 & 0 & 2 \end{vmatrix}$
= i (5(2) - 0(1)) - j (0(2) - (-1)1) + k (0(0) - (-1)5)
= 10 i - j + 5k.

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