WEBVTT
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Find, to the nearest degree, the measure of the angle between the plane two times the quantity 𝑥 minus three minus eight times the quantity 𝑦 plus two plus three 𝑧 equals four and the straight line 𝑥 minus three over six equals 𝑦 plus one over two equals 𝑧 minus one over four.
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Alright, so here we have a plane and a line.
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And we want to solve for the measure of the angle between these objects.
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We’re going to do this by discovering a vector that is perpendicular or normal to our plane and a vector that is parallel to our straight line.
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The reason we’re looking for vectors like these is that there’s a mathematical relationship for the sine of the angle between a plane and a straight line.
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And this equation involves a vector that is parallel to the line and one that is normal to the plane.
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And that’s what we’re after then.
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And let’s start with the given equation of a plane.
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We’re going to rearrange this expression a bit so that our plane will be in what’s called general form.
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If we multiply two and negative eight through these parentheses, then subtract four from both sides of the equation, and then collect all of our constants into one term, we now have our plane’s equation in what is known as general form.
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Written this way, we can recall that the components of a vector normal to this plane are given by the values by which we multiply 𝑥, 𝑦, and 𝑧, respectively.
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That is, a vector that’s normal to our plane has components two, negative eight, three.
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Next, we can look at the equation of our straight line in order to solve for a vector that is parallel to it.
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Our line’s equation is given to us in a form called symmetric form.
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We’re able to write the line’s equation this way because each one of these fractions is equal to the same scale factor.
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We can call that scale factor 𝑡.
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Knowing this, it’s possible to rewrite the equation of this line in what’s called parametric form.
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In this new form, we have one equation for 𝑥, one for 𝑦, and one for 𝑧.
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To write an equation for the 𝑥-values of the points on our line, we recognize that 𝑥 minus three over six is equal to 𝑡.
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That tells us that 𝑥 equals six times 𝑡 plus three.
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We write our 𝑦 equation based on the fact that 𝑦 plus one over two equals 𝑡.
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So 𝑦 must equal two times 𝑡 minus one.
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Likewise, 𝑧 minus one over four equaling 𝑡 tells us that 𝑧 equals four times 𝑡 plus one.
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Our line is now written in parametric form.
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And we’re now able to write all three of these equations in one summary equation.
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If we let the vector 𝐫 represent the equation of our line, then our parametric equations tell us that this line passes through the point three, negative one, one and is parallel to the vector six, two, four.
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This means we now know the components of a vector parallel to our straight line and the components of one normal to our plane.
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If we substitute these vectors into this equation for the sine of the angle between the line and the plane, then we get this expression.
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If we start evaluating this by computing the dot product in the numerator and squaring out the components of our vectors in the denominator and then continue to simplify, we get a simplified result of eight over the square root of 56 times 77.
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This fraction is equal to the sine of the angle we want to solve for.
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So to compute the angle itself, we’ll take the inverse sine of both sides.
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Calculating this expression, to the nearest degree, we get a result of seven degrees.
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This is the measure of the angle between the given plane and the given line.