H2 Maths 3.2 - Scalar and Vector Products Worked Example

Step 1 - Set up the problem

Let O be the origin, with a equals i plus two j minus two k, and b equals three i minus j plus k.

Step 1 - Set up the problem

We want to find the scalar product and the angle between them.

Step 1 - Set up the problem

Then we will find the vector product, and use it to compute the area of triangle OAB.

Step 2 - Compute the scalar product

The scalar product multiplies corresponding components and sums them.

Step 2 - Compute the scalar product

One times three is three, two times negative one is negative two, and negative two times one is negative two.

Step 2 - Compute the scalar product

Three minus two minus two equals negative one.

Step 3 - Find the angle between the vectors

Use the formula: a dot b equals the product of the magnitudes times cosine theta.

Step 3 - Find the angle between the vectors

The magnitude of a is the square root of one plus four plus four, which is exactly three.

Step 3 - Find the angle between the vectors

The magnitude of b is the square root of nine plus one plus one, which is root eleven.

Step 3 - Find the angle between the vectors

So cosine theta equals negative one over three root 11, giving about 95.8 degrees.

Step 3 - Find the angle between the vectors

The obtuse answer confirms the negative dot product.

Step 4 - Compute the vector product

Write the three-by-three determinant with i, j, k in the top row.

Step 4 - Compute the vector product

The i component is two times one minus negative two times negative one - that is two minus two, giving zero.

Step 4 - Compute the vector product

The j component is negative of one times one minus negative two times three - that is negative of one plus six, giving negative seven.

Step 4 - Compute the vector product

The k component is one times negative one minus two times three - that is negative one minus six, giving negative seven.

Step 5 - Area of triangle OAB and key pitfall

The magnitude of a cross b is root of zero plus 49 plus 49.

Step 5 - Area of triangle OAB and key pitfall

Root 98 simplifies to seven root two.

Step 5 - Area of triangle OAB and key pitfall

The cross product magnitude equals the parallelogram area, so the triangle area is half of seven root two.

Step 5 - Area of triangle OAB and key pitfall

Common pitfall: b cross a equals the negative of a cross b, so order always matters.