Question Video: Finding the Angle between Two Given Vectors | Nagwa Question Video: Finding the Angle between Two Given Vectors | Nagwa

# Question Video: Finding the Angle between Two Given Vectors Mathematics • Third Year of Secondary School

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Find the angle π between the vectors π = <5, 1, β2> and π = <4, β4, 3>. Give your answer correct to two decimal places.

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### Video Transcript

Find the angle π between the vectors π five, one, negative two and π four, negative four, three. Give your answer correct to two decimal places.

In this question, weβre asked to determine the angle π between two vectors, the vector π and the vector π, given in component form. We need to give our value of π correct to two decimal places. To help us answer this question, itβs worth recalling how we find the angle between two vectors. We recall if π is the angle between two vectors π and π, then the cos of angle π will be equal to the dot product of vectors π and π divided by the magnitude of vector π times the magnitude of vector π. And itβs worth pointing out the same is true in reverse. If π satisfies this equation, then we can say that π is an angle between vectors π and π.

However, by convention, when we say the angle between two vectors, we mean the smallest nonnegative angle between these two vectors. In this case, we can find this by taking the inverse cosine of both sides of the equation. What this really means is to find the angle between two vectors, we need to know their dot product and the magnitude of the two vectors π and π. Therefore, to find the angle π between our two vectors π and π, we need to find the dot product between π and π, the magnitude of vector π, and the magnitude of vector π. Letβs start by finding the dot product between vector π and vector π.

To do this, we need to recall to find the dot product between two vectors, we need to find the sum of the products of the corresponding components of the two vectors. In this case, thatβs five multiplied by four plus one multiplied by negative four plus negative two multiplied by three, which if we evaluate this expression, we get that itβs equal to 10. Next, we need to calculate the magnitude of vectors π and π. To do this, we recall the magnitude of a vector is equal to the square root of the sum of the squares of its components. In other words, the magnitude of vector π, π, π will be equal to the square root of π squared plus π squared plus π squared.

We can use this to find the magnitude of vector π. Thatβs the magnitude of the vector five, one, negative two. The magnitude of vector π will be the square root of the sum of the squares of its components. The magnitude of π is the square root of five squared plus one squared plus negative two all squared, which if we evaluate the expression inside our square root symbol, we get the square root of 30. We can then do exactly the same to find the magnitude of vector π. Itβs equal to the square root of four squared plus negative four all squared plus three squared, which we can simplify to give us the square root of 41. Weβre now ready to find an expression for our value of π.

First, we know since π is the angle between the vectors π and π, the cos of π will be equal to the dot product between vector π and vector π divided by the magnitude of vector π times the magnitude of vector π. We can then substitute the values we found for the dot product between vector π and vector π and the magnitude of vector π and vector π. We get the cos of π will be equal to 10 divided by root 30 multiplied by root 41. We can then solve for the value of π by taking the inverse cosine of both sides of this equation. Remember, this will give us the smallest nonnegative angle between the two vectors π and π. We get that π will be equal to the inverse cos of 10 divided by root 30 multiplied by root 41.

Finally, we can calculate this value in degrees. We get that π is 73.433 and this continues degrees. But remember, the question wants us to give our answer to two decimal places. To do this, we look at the third decimal place, which is three. Since this is less than five, this means we need to round down. And this gives us our final answer. The angle π between the vector π five, one, negative two and the vector π four, negative four, three to two decimal places is 73.43 degrees.

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