Scalar and vector products 285
is parallel to the vector 2i + 7j −4k. Determine
the point on the line corresponding to λ =2 in
the resulting equation.
⎡
⎣
r = (5 + 2λ)i + (7λ − 2)j
+ (3 − 4λ)k;
r = 9i + 12j − 5k
⎤
⎦
2. Express the vector equation of the line in
problem 1 in standard Cartesian form.
x − 5
2
=
y + 2
7
=
3 − z
4
= λ
In problems 3 and 4, express the given straight line
equations in vector form.
3.
3x − 1
4
=
5y + 1
2
=
4 − z
3
r =
1
3 (1 + 4λ)i +
1
5 (2λ − 1)j
+ (4 − 3λ)k
4. 2x + 1 =
1 −4y
5
=
3z −1
4
r =
1
2 (λ − 1)i +
1
4 (1 − 5λ)j
+
1
3 (1 + 4λ)k
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