Question Video: Finding the Value of a Coefficient Using the Angle between a Line and Plane | Nagwa Question Video: Finding the Value of a Coefficient Using the Angle between a Line and Plane | Nagwa

# Question Video: Finding the Value of a Coefficient Using the Angle between a Line and Plane Mathematics • Third Year of Secondary School

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If the angle between the plane 3π₯ + ππ¦ β 5π§ = 2 and the straight line π₯ + 2 = (3 β π¦)/2 = (π§ β 1)/2 is 45Β°, find the value of the positive constant π given that 0 < π < 10.

06:20

### Video Transcript

If the angle between the plane three π₯ plus ππ¦ minus five π§ is equal to two and the straight line π₯ plus two is equal to three minus π¦ over two is equal to π§ minus one over two is 45 degrees, find the value of the positive constant π given that π is greater than zero and less than 10.

In this question, weβre given the equation of a plane and the equation of a straight line. And our equation of a plane involves an unknown value of π. We need to determine the value of π by using the fact itβs a positive constant which is less than 10 and the angle between the line and the plane is 45 degrees. And to find this value of π, letβs start by recalling how we find the angle between a line and a plane. We know if the vector π is a direction vector of a line πΏ and the vector π§ is a normal vector to a plane π, then the acute angle π between the line and the plane will satisfy the equation sin of π is equal to the absolute value of the dot product of π§ and π divided by the magnitude of vector π§ multiplied by the magnitude of vector π.

And there is one small thing worth noting. Since weβre taking the dot product of vector π§ and π, we are working in three dimensions for both our line and our plane. Weβre told in the question the angle between the given line and plane is 45 degrees, so our value of π is 45. We can find vectors π and π§ by looking at the equations of the line in the plane. Letβs start by finding the direction vector of the line. To find the direction vector of this line, we recall a line in the form π₯ minus π₯ one over π is equal to π¦ minus π¦ one over π is equal to π§ minus π§ one over π will have a direction vector of π, π, π. And our line is almost given in this form. Thereβs just a few changes we need to make.

First, on the leftmost part of this equation, weβre going to divide through by one. Next, in the numerator of the middle part of this equation, we donβt have π¦ in the numerator; we have negative π¦. So weβre going to need to multiply the numerator and denominator by negative one. If we distributed the parentheses, this would now be in the required form, and we can find the direction vector of this line. The direction vector π is the vector one, negative two, two.

Letβs now find the normal vector to the plane. We can do this by recalling the plane ππ₯ plus ππ¦ plus ππ§ is equal to zero 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 and is the vector three, π, negative five. We could now substitute our value for π and our vectors π§ and π into our equation. However, itβs easier to evaluate the numerator and denominator of the right-hand side separately.

So letβs start by clearing some space and then calculating the dot product between vectors π§ and π. We can do this by recalling to evaluate the dot product of two vectors of equal dimension, we just need to find the sum of the products of the corresponding components of the vectors. Applying this, we get one times three plus negative two multiplied by π plus two times negative five, which if we evaluate gives us negative two π minus seven.

Letβs now determine the magnitudes of the two vectors. Letβs start with the magnitude of vector π§. We can find this by recalling the magnitude of a vector is the square root of the sum of the squares of its components. So we get the square root of three squared plus π squared plus negative five squared, which if we evaluate gives us the square root of 34 plus π squared. We can do the same to find the magnitude of vector π. Itβs the square root of one squared plus negative two squared plus two squared, which we can evaluate is the square root of nine, which is just equal to three.

Letβs now clear some space and substitute all of this information into our equation involving the acute angle between the line and plane. This then gives us that the sin of 45 degrees must be equal to the absolute value of negative two π minus seven divided by three times the square root of 34 plus π squared. We now want to solve this equation for π. We can do this by first noting the sin of 45 degrees is equal to root two over two. And we can also multiply both sides of the equation through by three root 34 plus π squared. This then gives us root two over two multiplied by three times the square root of 34 plus π squared is equal to the absolute value of negative two π minus seven.

And remember, weβre told in the question that the value of π is positive. So negative two multiplied by π is a negative multiplied by a positive. Itβs a negative number. We then subtract seven, so this is negative. And when we take the absolute value of a negative number, we just multiply it by negative one. So the right-hand side of this equation simplifies to give us two π plus seven.

We can simplify the left-hand side of this equation. We have root two multiplied by root 34 π squared. We can simplify this as the square root of two times 34 plus π squared. This gives us the square root of 68 plus two π squared. We need to multiply this by three over two. However, weβll just multiply this by three and then multiply both sides of our equation through by two. This gives us that three root 68 plus two π squared is equal to two times two π plus seven. We can distribute the two over the parentheses on the right-hand side of our equation. This gives us four π plus 14.

Now, to solve our equation for π, weβre going to need to square both sides of our equation. On the left-hand side, three squared is nine. And then to square a square root, we can just remove the square root symbol. We get nine times 68 plus two π squared. And on the right-hand side of this equation, we have the square of a binomial. We can solve this by using the FOIL method or binomial expansion. Either way, we get 16π squared plus 112π plus 196.

Now, weβre going to distribute nine over our parentheses and then rearrange to get a quadratic in π. Doing this, we get two π squared minus 112π plus 416 is equal to zero. And we can simplify this slightly by noting all three of our terms share a factor of two. So we can divide our equation through by two to get π squared minus 56π plus 280 is equal to zero. We can then solve this by using the quadratic formula, or we can look for two numbers which multiply to give 208 and add to give negative 56.

Using either method, we can find two solutions. Either our value of π is equal to four or our value of π is 52. However, weβre told in the question our value of π is positive and less than 10. So only one of these is a valid value for π. Therefore, we were able to show there was only one possible value of the positive constant π between zero and 10. And that value of π was four.

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