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Find the point of intersection of the line π₯ minus six over four is equal to π¦ plus three is equal to π§ with the plane π₯ plus three π¦ plus two π§ minus six is equal to zero.
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The point of intersection π₯, π¦, π§ of a line and a plane is given by the unique solution of the system of equations consisting of the line and the plane equations.
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There are various methods available to solve such systems of equations.
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And for this example, weβll solve them algebraically.
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The equation of the line that weβre given is effectively two equations since we have two equalities.
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And we start by writing the equation of the line as two distinct equations involving π§.
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The first of these is π§ is equal to π₯ minus six over four, and the second is π§ is π¦ plus three.
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Multiplying both sides by four in the first equation gives us four π§ is π₯ minus six.
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And adding six to both sides, we then have π₯ in terms of π§; thatβs equal to four π§ plus six.
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And in our second equation, we simply subtract three from both sides to give π¦ is equal to π§ minus three.
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So now, together with our plane equation, we have three equations and three unknowns.
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And as long as the line and the plane are not parallel or coplanar, there should be one unique solution to these equations, and that is our point of intersection.
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The equation of the plane is π₯ plus three π¦ plus two π§ minus six is equal to zero.
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And to solve for a point of intersection, we substitute our π₯- and π¦-expressions just obtained into the equation of the plane.
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This gives us four π§ plus six, thatβs π₯, plus three multiplied by π§ minus three, thatβs three times π¦, plus two π§ minus six is equal to zero.
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That is four π§ plus six plus three π§ minus nine plus two π§ minus six is equal to zero.
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And collecting like terms, that is nine π§ minus nine is equal to zero.
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And then solving for π§, thatβs nine π§ is equal to nine so that π§ is equal to one.
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So now making a note of this and making some room, now remembering that the equation of the line tells us that π₯ is equal to four π§ plus six and π¦ is π§ minus three.
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Substituting in our value of π§ is equal to one, we have π₯ is equal to four multiplied by one plus six.
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That is π₯ is equal to 10.
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And substituting π§ is equal to one into π¦, we have π¦ is equal to one minus three.
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That is π¦ is negative two.
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Our point of intersection of the given line and plane is therefore 10, negative two, one.
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We can check that all of these values are correct by substituting back into the equations.
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Into our plane equation, we have 10 plus three times negative two plus two times one minus six.
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That is 10 minus six plus two minus six, and that is actually zero.
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And next into our line equation, we have 10 minus six over four is equal to negative two plus three is equal to one, and these are indeed all equal to one.
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Our point of intersection is therefore 10, negative two, one.