Straight line through A (with position vector a) and parallel to a vector b. Let P be a general point on L, then \overrightarrow{OP} =\overrightarrow{ OA} +\overrightarrow{ AP} = r = a +\overrightarrow{ AP}. Since \overrightarrow{AP} is parallel to b , hence \overrightarrow{AP} = λb (for some scalar λ), λ may be positive or negative. Thus r = a + λb. This is the vector equation of a straight line.
If r = xi + yj + zk, and a = {a}_{1}i + {a}_{2}j + {a}_{3}k, b = {b}_{1}i + {b}_{2}j + {b}_{3}k the equation
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r = a + λb,
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gives xi + yj + zk = ({a}_{1} + λ{b}_{1})i + ({a}_{2} + λ{b}_{2})j + ({a}_{3} + λ{b}_{3})k. Equality of the vectors gives 3 scalar equations, x = {a}_{1} + λ{b}_{1} or {(x−{a}_{1})\over {b}_{1}} = λ, y = {a}_{2} + λ{b}_{2} or (y − {a}_{2})∕{b}_{2} = λ and z = {a}_{3} + λ{b}_{3} or (z − {a}_{3})∕{b}_{3} = λ. Since −∞ < λ < ∞, (for different points on L), we find that these three scalar equations give the Cartesian equations of L as
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{x − {a}_{1}\over
{b}_{1}} = {y − {a}_{2}\over
{b}_{2}} = {z − {a}_{3}\over
{b}_{3}} = λ
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This is called the standard or canonical form.
Example 3.14:
Find the position vector of a point on a straight line L and a vector along L whose Cartesian equations are
{3x+1\over
2} = {y−7\over
3} = {−2z+1\over
4} .
the standard form of L is
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{x + {1\over
3}\over
{2\over
3}} = {y − 7\over
3} = {z −{1\over
2}\over
−2}
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Point A: \Big (−{1\over 3},7, {1\over 2}\Big ), position vector of A: −{1\over 3}i + 7j + {1\over 2}k. b = {2\over 3}i + 3j − 2k (parallel to L)
Example 3.15:
Example: Find the Cartesian equations of a straight line L through the point
a = i− 2j + k in the direction
of the vector b = −2j + 3k.
L: r = a + λb
gives xi + yj + zk = i− 2j + k + λ(0.i− 2j + 3k).
This gives the following Cartesian equations of L:
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{x − 1\over
0} = {y + 2\over
−2} = {z − 1\over
3} (= λ)
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