Video Transcript
If vector 𝐀 is equal to two 𝐢 plus five 𝐤 and vector 𝐁 is equal to four 𝐢 plus three 𝐣 plus 𝐤, find the measure of the angle between the two vectors rounded to the nearest hundredth.
We recall that the cos of the angle between two vectors 𝜃 is equal to the dot product of the two vectors divided by the product of the magnitude of the two vectors. In this question, we begin by calculating the dot product. We do this by multiplying the coefficients of 𝐢, the coefficients of 𝐣, and the coefficients of 𝐤. We then find the sum of these three numbers.
Two multiplied by four is equal to eight. There is no 𝐣-component in vector 𝐀. Therefore, the coefficient is zero. Zero multiplied by three equals zero. Five multiplied by one is equal to five. Eight plus zero plus five is equal to 13. Therefore, the dot product of vectors 𝐀 and 𝐁 is 13.
The magnitude of any vector is calculated by square rooting 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 𝐀 is, therefore, equal to the square root of two squared plus zero squared plus five squared. This is equal to the square root of 29. We repeat this process to find the magnitude of vector 𝐁. This is equal to the square root of four squared plus three squared plus one squared. This is equal to the square root of 26. We can now substitute our three values into the formula.
The cos of 𝜃 is equal to 13 divided by the square root of 29 multiplied by the square root of 26. Taking the inverse cos of both sides of this equation gives us a value of 𝜃 equal to cos to the minus one of 13 divided by the square root of 29 multiplied by the square root of 26. This is equal to 61.7426 and so on. We need to round to the nearest hundredth. So, our deciding number is the two. This means we’ll round down. The measure of the angle between the two vectors is 61.74 degrees.
