A line in space has the vector equation shown here. Fill in the blanks to find the position vector to the fixed point on the line that appears in the equation:
vector r=(5+(6/11)d)vector i + (3+(9/11)d)vector j + (7+(2/11)d)vector k
vector v= _____ vector i + _____ vector j + ____ vector k
Write the coordinates of the point on the line that is 11 units in the positive d-direction away from the fixed point in this problem.
What is the "positive d-direction"? d is not a direction, it is a parameter for r(d).
If d is a parameter, any point involving d is not fixed.
Is the fixed point 5i + 3j + 7k and you want r(11)?
Yeah, sorry about that. There's a diagram to it and I wasn't sure how else to explain it.
Is it just ((6/11), (9/11), (2/11)) ?
To find the position vector to the fixed point on the line, we need to determine the coefficients of the unit vectors (i, j, k) in the vector equation r = (5 + (6/11)d)i + (3 + (9/11)d)j + (7 + (2/11)d)k.
Comparing the coefficients, we can see that:
- The coefficient of i is (5 + (6/11)d).
- The coefficient of j is (3 + (9/11)d).
- The coefficient of k is (7 + (2/11)d).
Therefore, the vector v representing the position vector is:
v = (5 + (6/11)d)i + (3 + (9/11)d)j + (7 + (2/11)d)k.
To find the coordinates of the point on the line that is 11 units in the positive d-direction away from the fixed point, we substitute d with d + 11 in the vector equation:
v = (5 + (6/11)(d + 11))i + (3 + (9/11)(d + 11))j + (7 + (2/11)(d + 11))k
Simplifying the equation, we have:
v = (5 + 6 + 66/11)i + (3 + 9 + 99/11)j + (7 + 2 + 22/11)k
= (11 + 66/11)i + (12 + 99/11)j + (9 + 22/11)k
Therefore, the coordinates of the point on the line that is 11 units in the positive d-direction away from the fixed point are:
(11 + 66/11, 12 + 99/11, 9 + 22/11)