Find the angle 𝜃 between the vectors 𝐯 two, one, four and 𝐰 one, negative two, zero.
We recall that the cos of angle 𝜃, the angle between two vectors, is equal to the dot product of the two vectors, 𝐚 dot 𝐛, divided by the product of the magnitude or modulus of the two vectors. In this question, vector 𝐯 is equal to two 𝐢 plus 𝐣 plus four 𝐤. Vector 𝐰 is equal to 𝐢 minus two 𝐣 plus zero 𝐤. This can be simplified to 𝐢 minus two 𝐣. The dot product of these two vectors can be found by multiplying the coefficients of 𝐢, the coefficients of 𝐣, and the coefficients of 𝐤. We then calculate the sum of these three answers.
Two multiplied by one is equal to two. One multiplied by negative two is negative two. Finally, four multiplied by zero is equal to zero. Two minus two plus zero is equal to zero. Therefore, the dot product of vectors 𝐯 and 𝐰 is zero. The magnitude of vector 𝐚 is equal to the square root of 𝑥 squared plus 𝑦 squared plus 𝑧 squared, where 𝑥, 𝑦, and 𝑧 are the coefficients of 𝐢, 𝐣, and 𝐤, respectively. The magnitude of vector 𝐯 is therefore equal to the square root of two squared plus one squared plus four squared. This is equal to the square root of 21.
The magnitude of 𝐰 can be calculated in the same way. One squared plus negative two squared plus zero squared. This is equal to the square root of five. We can now substitute these three values into the formula. The cos of 𝜃 is equal to zero divided by the square root of 21 multiplied by the square root of five. Zero divided by any number is equal to zero. Therefore, the cos of 𝜃 equals zero. Taking the inverse cos of both sides of this equation gives us 𝜃 is equal to cos to the minus one or inverse cos of zero. This is equal to 90 degrees. The angle between the two vectors 𝐯 and 𝐰 is 90 degrees.