Question Video: Finding the Angle between Planes | Nagwa Question Video: Finding the Angle between Planes | Nagwa

# Question Video: Finding the Angle between Planes Mathematics • Third Year of Secondary School

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Find, to the nearest degree, the measure of the angle between the planes 3π₯ β 2π¦ + 3π§ = 7 and 2(π₯ β 1) + 3(π¦ β 4) + 4(π§ + 5) = 0.

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### Video Transcript

Find, to the nearest degree, the measure of the angle between the planes three π₯ minus two π¦ plus three π§ is equal to seven and two times π₯ minus one plus three times π¦ minus four plus four times π§ plus five is equal to zero.

In this question, weβre asked to find the measure of the angle between two given planes, and we need to find this angle to the nearest degree. To do this, letβs start by recalling how we find the measure of the angle between two given planes. We know if we have two planes with normal vectors π§ sub one and π§ sub two, then the acute angle π between the two planes will satisfy the equation cos of π is equal to the absolute value of the dot product of vectors π§ sub one and π§ sub two divided by the magnitude of π§ sub one multiplied by the magnitude of π§ sub two. So, we can find an equation involving the angle between the two planes if we can find a normal vector to each plane.

We can find the normal vector to both of our planes by recalling the plane ππ₯ plus ππ¦ plus ππ§ is equal to π will have a normal vector of π, π, π. In other words, the components of the normal vector to the plane are the coefficients of the variables. We can directly find the normal vector of the first plane by just looking at the coefficients of the variables. We get the normal vector for the first plane π§ sub one is equal to the vector three, negative two, three.

We now want to find the normal vector of the second plane. And we can do this by first distributing all of our coefficients over the parentheses. However, we donβt need to do this in full; weβre only interested in the coefficients of our variables. So, we only need to calculate these terms. We get two π₯, three π¦, and four π§. And then taking these coefficients as the components of our vector gives us the normal vectors to this plane π§ sub two is the vector two, three, four.

We could now substitute these vectors into our equation involving the angle π. However, itβs usually easier to determine the numerator and denominator of the right-hand side of this equation separately. Letβs first find the dot product of the two vectors. We want to find the dot product of the vector three, negative two, three and the vector two, three, four. And we can do this by recalling to find the dot product of two vectors of equal dimension, we just need to find the sum of the products of the corresponding components. In this case, thatβs three times two plus negative two multiplied by three plus three times four, which is equal to 12. Remember, in the right-hand side of our equation, we want to find the absolute value of this dot product. Well, the absolute value of 12 is just equal to 12.

We now want to evaluate the denominator of the right-hand side of our equation. To do this, we need to find the magnitude of our two normal vectors. Letβs start with the magnitude of π§ sub one. Remember, the magnitude of a vector is the square root of the sum of the squares of its components. So, the magnitude of vector π§ sub one is the square root of three squared plus negative two squared plus three squared, which we can then calculate is the square root of 22. We can follow the same process to find the magnitude of vector π§ sub two. Itβs equal to the square root of two squared plus three squared plus four squared, which we can evaluate is root 29.

Weβre now ready to substitute these values into our equation. This gives us that the cos of π is equal to 12 divided by root 22 times root 29. We could simplify the right-hand side of this equation. However, itβs not necessary since we now need to take the inverse cosine of both sides of the equation anyway. This gives us that π is equal to the inverse cos of 12 divided by root 22 times root 29.

And if we input this expression into our calculator, where we make sure itβs set to degrees mode, we get 61.63 and this expansion continues degrees. But remember, the question wants us to determine this angle to the nearest degree. So, we need to look at the first decimal digit, which is six. This is greater than or equal to five. So, we need to round this value up, which then gives us our final answer.

The measure of the angle between the planes three π₯ minus two π¦ plus three π§ is equal to seven and two times π₯ minus one plus three times π¦ minus four plus four times π§ plus five is equal to zero to the nearest degree is 62 degrees.

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