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VECTORS AND THE GEOMETRY OF SPACE A line in the xy-plane is determined when a point on the line and the direction of the line (its slope or angle of inclination) are given. The equation of the line can then be written using the point-slope form.

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EQUATIONS OF LINES A line L in three-dimensional (3-D) space is determined when we know: A point P0(x0, y0, z0) on L The direction of L

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EQUATIONS OF LINES In three dimensions, the direction of a line is conveniently described by a vector.

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EQUATIONS OF LINES So, we let v be a vector parallel to L. Let P(x, y, z) be an arbitrary point on L. Let r0 and r be the position vectors of P0 and P. That is, they have representations and .

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EQUATIONS OF LINES If a is the vector with representation , then the Triangle Law for vector addition gives: r = r0 + a

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EQUATIONS OF LINES However, since a and v are parallel vectors, there is a scalar t such that a = tv

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VECTOR EQUATION OF A LINE Thus, r = r0 + t v This is a vector equation of L.

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VECTOR EQUATION Each value of the parameter t gives the position vector r of a point on L. That is, as t varies, the line is traced out by the tip of the vector r.

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VECTOR EQUATION Positive values of t correspond to points on L that lie on one side of P0. Negative values correspond to points that lie on the other side.

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VECTOR EQUATION If the vector v that gives the direction of the line L is written in component form as v = , then we have: tv =

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VECTOR EQUATION We can also write: r = and r0 = So, vector Equation 1 becomes: =

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VECTOR EQUATION Two vectors are equal if and only if corresponding components are equal. Hence, we have the following three scalar equations.

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SCALAR EQUATIONS OF A LINE x = x0 + at y = y0 + bt z = z0 + ct Where, t

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PARAMETRIC EQUATIONS These equations are called parametric equations of the line L through the point P0(x0, y0, z0) and parallel to the vector v = . Each value of the parameter t gives a point (x, y, z) on L.

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EQUATIONS OF LINES Find a vector equation and parametric equations for the line that passes through the point (5, 1, 3) and is parallel to the vector i + 4 j – 2 k. Find two other points on the line.

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EQUATIONS OF LINES Here, r0 = = 5 i + j + 3 k and v = i + 4 j – 2 k So, vector Equation 1 becomes: r = (5 i + j + 3 k) + t(i + 4 j – 2 k) or r = (5 + t) i + (1 + 4t) j + (3 – 2t) k

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EQUATIONS OF LINES Parametric equations are: x = 5 + t y = 1 + 4t z = 3 – 2t

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EQUATIONS OF LINES Choosing the parameter value t = 1 gives x = 6, y = 5, and z = 1. So, (6, 5, 1) is a point on the line. Similarly, t = –1 gives the point (4, –3, 5).

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EQUATIONS OF LINES The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change.

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EQUATIONS OF LINES For instance, if, instead of (5, 1, 3), we choose the point (6, 5, 1) in Example 1, the parametric equations of the line become: x = 6 + t y = 5 + 4t z = 1 – 2t

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EQUATIONS OF LINES Alternatively, if we stay with the point (5, 1, 3) but choose the parallel vector 2 i + 8 j – 4 k, we arrive at: x = 5 + 2t y = 1 + 8t z = 3 – 4t

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DIRECTION NUMBERS In general, if a vector v = is used to describe the direction of a line L, then the numbers a, b, and c are called direction numbers of L.

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DIRECTION NUMBERS Any vector parallel to v could also be used. Thus, we see that any three numbers proportional to a, b, and c could also be used as a set of direction numbers for L.

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EQUATIONS OF LINES Another way of describing a line L is to eliminate the parameter t from Equations 2. If none of a, b, or c is 0, we can solve each of these equations for t, equate the results, and obtain the following equations.

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SYMMETRIC EQUATIONS These equations are called symmetric equations of L.

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SYMMETRIC EQUATIONS Notice that the numbers a, b, and c that appear in the denominators of Equations 3 are direction numbers of L. That is, they are components of a vector parallel to L.

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SYMMETRIC EQUATIONS If one of a, b, or c is 0, we can still eliminate t. For instance, if a = 0, we could write the equations of L as: This means that L lies in the vertical plane x = x0.

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EQUATIONS OF LINES Find parametric equations and symmetric equations of the line that passes through the points A(2, 4, –3) and B(3, –1, 1). At what point does this line intersect the xy-plane?

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EQUATIONS OF LINES We are not explicitly given a vector parallel to the line. However, observe that the vector v with representation is parallel to the line and v = =

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EQUATIONS OF LINES Thus, direction numbers are: a = 1, b = –5, c = 4

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EQUATIONS OF LINES Taking the point (2, 4, –3) as P0, we see that: Parametric Equations 2 are: x = 2 + t y = 4 – 5t z = –3 + 4t Symmetric Equations 3 are:

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EQUATIONS OF LINES The line intersects the xy-plane when z = 0. So, we put z = 0 in the symmetric equations and obtain: This gives x = and y = .

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EQUATIONS OF LINES The line intersects the xy-plane at the point

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EQUATIONS OF LINES In general, the procedure of Example 2 shows that direction numbers of the line L through the points P0(x0, y0, z0) and P1(x1, y1, z1) are: x1 – x0 y1 – y0 z1 – z0 So, symmetric equations of L are:

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EQUATIONS OF LINE SEGMENTS Often, we need a description, not of an entire line, but of just a line segment. How, for instance, could we describe the line segment AB in Example 2?

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EQUATIONS OF LINE SEGMENTS If we put t = 0 in the parametric equations in Example 2 a, we get the point (2, 4, –3). If we put t = 1, we get (3, –1, 1).

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EQUATIONS OF LINE SEGMENTS So, the line segment AB is described by either: The parametric equations x = 2 + t y = 4 – 5t z = –3 + 4t where 0 ≤ t ≤ 1 The corresponding vector equation r(t) = where 0 ≤ t ≤ 1

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EQUATIONS OF LINE SEGMENTS In general, we know from Equation 1 that the vector equation of a line through the (tip of the) vector r0 in the direction of a vector v is: r = r0 + t v

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EQUATIONS OF LINE SEGMENTS If the line also passes through (the tip of) r1, then we can take v = r1 – r0. So, its vector equation is: r = r0 + t(r1 – r0) = (1 – t)r0 + t r1 The line segment from r0 to r1 is given by the parameter interval 0 ≤ t ≤ 1.

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EQUATIONS OF LINE SEGMENTS The line segment from r0 to r1 is given by the vector equation r(t) = (1 – t)r0 + t r1 where 0 ≤ t ≤ 1

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EQUATIONS OF LINE SEGMENTS Show that the lines L1 and L2 with parametric equations x = 1 + t y = –2 + 3t z = 4 – t x = 2s y = 3 + s z = –3 + 4s are skew lines. That is, they do not intersect and are not parallel, and therefore do not lie in the same plane.

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EQUATIONS OF LINE SEGMENTS The lines are not parallel because the corresponding vectors and are not parallel. Their components are not proportional.

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EQUATIONS OF LINE SEGMENTS If L1 and L2 had a point of intersection, there would be values of t and s such that 1 + t = 2s –2 + 3t = 3 + s 4 – t = –3 + 4s

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EQUATIONS OF LINE SEGMENTS However, if we solve the first two equations, we get: t = and s = These values don’t satisfy the third equation.

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EQUATIONS OF LINE SEGMENTS Thus, there are no values of t and s that satisfy the three equations. So, L1 and L2 do not intersect.

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EQUATIONS OF LINE SEGMENTS Hence, L1 and L2 are skew lines.

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PLANES Although a line in space is determined by a point and a direction, a plane in space is more difficult to describe. A single vector parallel to a plane is not enough to convey the ‘direction’ of the plane.

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PLANES However, a vector perpendicular to the plane does completely specify its direction.

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PLANES Thus, a plane in space is determined by: A point P0(x0, y0, z0) in the plane A vector n that is orthogonal to the plane

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NORMAL VECTOR This orthogonal vector n is called a normal vector.

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PLANES Let P(x, y, z) be an arbitrary point in the plane. Let r0 and r1 be the position vectors of P0 and P. Then, the vector r – r0 is represented by

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PLANES The normal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to r – r0.

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EQUATIONS OF PLANES Thus, we have: n . (r – r0) = 0

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EQUATIONS OF PLANES That can also be written as: n . r = n . r0

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VECTOR EQUATION Either Equation 5 or Equation 6 is called a vector equation of the plane.

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EQUATIONS OF PLANES To obtain a scalar equation for the plane, we write: n = r = r0 =

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EQUATIONS OF PLANES Then, the vector Equation 5 becomes: . = 0

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SCALAR EQUATION That can also be written as: a(x – x0) + b(y – y0) + c(z – z0) = 0 This equation is the scalar equation of the plane through P0(x0, y0, z0) with normal vector n = .

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EQUATIONS OF PLANES Find an equation of the plane through the point (2, 4, –1) with normal vector n = . Find the intercepts and sketch the plane.

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EQUATIONS OF PLANES In Equation 7, putting a = 2, b = 3, c = 4, x0 = 2, y0 = 4, z0 = –1, we see that an equation of the plane is: 2(x – 2) + 3(y – 4) + 4(z + 1) = 0 or 2x + 3y + 4z = 12

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EQUATIONS OF PLANES To find the x-intercept, we set y = z = 0 in the equation, and obtain x = 6. Similarly, the y-intercept is 4 and the z-intercept is 3.

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EQUATIONS OF PLANES This enables us to sketch the portion of the plane that lies in the first octant.

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EQUATIONS OF PLANES By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation of a plane as follows.

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LINEAR EQUATION ax + by + cz + d = 0 where d = –(ax0 + by0 + cz0) This is called a linear equation in x, y, and z.

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LINEAR EQUATION Conversely, it can be shown that, if a, b, and c are not all 0, then the linear Equation 8 represents a plane with normal vector . See Exercise 77.

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EQUATIONS OF PLANES Find an equation of the plane that passes through the points P(1, 3, 2), Q(3, –1, 6), R(5, 2, 0)

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EQUATIONS OF PLANES The vectors a and b corresponding to and are: a = b =

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EQUATIONS OF PLANES Since both a and b lie in the plane, their cross product a x b is orthogonal to the plane and can be taken as the normal vector.

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EQUATIONS OF PLANES Thus,

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EQUATIONS OF PLANES With the point P(1, 2, 3) and the normal vector n, an equation of the plane is: 12(x – 1) + 20(y – 3) + 14(z – 2) = 0 or 6x + 10y + 7z = 50

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EQUATIONS OF PLANES Find the point at which the line with parametric equations x = 2 + 3t y = –4t z = 5 + t intersects the plane 4x + 5y – 2z = 18

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EQUATIONS OF PLANES We substitute the expressions for x, y, and z from the parametric equations into the equation of the plane: 4(2 + 3t) + 5(–4t) – 2(5 + t) = 18

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EQUATIONS OF PLANES That simplifies to –10t = 20. Hence, t = –2. Therefore, the point of intersection occurs when the parameter value is t = –2.

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EQUATIONS OF PLANES Then, x = 2 + 3(–2) = –4 y = –4(–2) = 8 z = 5 – 2 = 3 So, the point of intersection is (–4, 8, 3).

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PARALLEL PLANES Two planes are parallel if their normal vectors are parallel.

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PARALLEL PLANES For instance, the planes x + 2y – 3z = 4 and 2x + 4y – 6z = 3 are parallel because: Their normal vectors are n1 = and n2 = and n2 = 2n1.

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NONPARALLEL PLANES If two planes are not parallel, then They intersect in a straight line. The angle between the two planes is defined as the acute angle between their normal vectors.

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EQUATIONS OF PLANES Find the angle between the planes x + y + z = 1 and x – 2y + 3z = 1 Find symmetric equations for the line of intersection L of these two planes.

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EQUATIONS OF PLANES The normal vectors of these planes are: n1 = n2 =

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EQUATIONS OF PLANES So, if θ is the angle between the planes, Corollary 6 in Section 12.3 gives:

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EQUATIONS OF PLANES We first need to find a point on L. For instance, we can find the point where the line intersects the xy-plane by setting z = 0 in the equations of both planes. This gives the equations x + y = 1 and x – 2y = 1 whose solution is x = 1, y = 0. So, the point (1, 0, 0) lies on L.

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EQUATIONS OF PLANES As L lies in both planes, it is perpendicular to both the normal vectors. Thus, a vector v parallel to L is given by the cross product

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EQUATIONS OF PLANES So, the symmetric equations of L can be written as:

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NOTE A linear equation in x, y, and z represents a plane. Also, two nonparallel planes intersect in a line. It follows that two linear equations can represent a line.

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NOTE The points (x, y, z) that satisfy both a1x + b1y + c1z + d1 = 0 and a2x + b2y + c2z + d2 = 0 lie on both of these planes. So, the pair of linear equations represents the line of intersection of the planes (if they are not parallel).

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NOTE For instance, in Example 7, the line L was given as the line of intersection of the planes x + y + z = 1 and x – 2y + 3z = 1

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NOTE The symmetric equations that we found for L could be written as: This is again a pair of linear equations.

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NOTE They exhibit L as the line of intersection of the planes (x – 1)/5 = y/(–2) and y/(–2) = z/(–3)

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NOTE In general, when we write the equations of a line in the symmetric form we can regard the line as the line of intersection of the two planes

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EQUATIONS OF PLANES Find a formula for the distance D from a point P1(x1, y1, z1) to the plane ax + by + cz + d = 0.

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EQUATIONS OF PLANES Let P0(x0, y0, z0) be any point in the plane. Let b be the vector corresponding to . Then, b =

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EQUATIONS OF PLANES You can see that the distance D from P1 to the plane is equal to the absolute value of the scalar projection of b onto the normal vector n = .

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EQUATIONS OF PLANES Thus,

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EQUATIONS OF PLANES Since P0 lies in the plane, its coordinates satisfy the equation of the plane. Thus, we have ax0 + by0 + cz0 + d = 0.

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EQUATIONS OF PLANES Hence, the formula for D can be written as:

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EQUATIONS OF PLANES Find the distance between the parallel planes 10x + 2y – 2z = 5 and 5x + y – z = 1

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EQUATIONS OF PLANES First, we note that the planes are parallel because their normal vectors and are parallel.

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EQUATIONS OF PLANES To find the distance D between the planes, we choose any point on one plane and calculate its distance to the other plane. In particular, if we put y = z =0 in the equation of the first plane, we get 10x = 5. So, (½, 0, 0) is a point in this plane.

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EQUATIONS OF PLANES By Formula 9, the distance between (½, 0, 0) and the plane 5x + y – z – 1 = 0 is: So, the distance between the planes is .

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EQUATIONS OF PLANES In Example 3, we showed that the lines L1: x = 1 + t y = –2 + 3t z = 4 – t L2: x = 2s y = 3 + s z = –3 + 4s are skew. Find the distance between them.

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EQUATIONS OF PLANES Since the two lines L1 and L2 are skew, they can be viewed as lying on two parallel planes P1 and P2. The distance between L1 and L2 is the same as the distance between P1 and P2. This can be computed as in Example 9.

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EQUATIONS OF PLANES The common normal vector to both planes must be orthogonal to both v1 = (direction of L1) v2 = (direction of L2)

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EQUATIONS OF PLANES So, a normal vector is:

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EQUATIONS OF PLANES If we put s = 0 in the equations of L2, we get the point (0, 3, –3) on L2. So, an equation for P2 is: 13(x – 0) – 6(y – 3) – 5(z + 3) = 0 or 13x – 6y – 5z + 3 = 0