If vector a=2i-j+k, b=i-3j+k,c= -2i+j-3k,d=3i+2j+5k find the p,q,r such that d=pa+qb+rc.
Please help me how to do it...!!!
I will use the notation
a = (2,-1,1) , b= (1,-3,1) , c=(-2,1,-3) and d=(3,2,5)
so
2p + q - 2r = 3
-p - 3q + r = 2
-2p + q - 3r = 5
add 1st and 3rd:
2q - 5r = 8 **
double the 2nd , then add to the first:
-5q = 7
q = -7/5
into **
-14/5 - 5r = 8
-5r = 54/5
r = -54/25
into the 1st:
2p - 7/5 +108/25 = 3
2p = 2/25
p = 1/25
Wolfram confirms:
http://www.wolframalpha.com/input/?i=solve+2p+%2B+q+-+2r+%3D+3,+-p+-+3q+%2B+r+%3D+2,+-2p+%2B+q+-+3r+%3D+5
oops, just noticed my third equation is wrong
Here we go again:
I will use the notation
a = (2,-1,1) , b= (1,-3,1) , c=(-2,1,-3) and d=(3,2,5)
so
2p + q - 2r = 3
-p - 3q + r = 2
p + q - 3r = 5
add the 2nd and 3rd:
-2q - 2r = 7
2q +2r = -7 **
double the 2nd , then add to the first:
-5q = 7
q = -7/5
into **
-14/5 + 2r = -7
2r = -21/5
r = -21/10
into the 1st:
2p -7/5 + 42/10 = 3
2p = 1/5
p = 1/10
check in original setup
(1/10)(2,-1,1) + (-7/5)(1,-3,1) + (-21/10)(-2,1,-3)
= (2/10,-1/10,-1/10) + (-7/5,21/5,-7/5) + (42/10,-21/10,63/10)
= (3, 2 , 5)
@Reiny...Thank you :)
To find the values of p, q, and r such that d = pa + qb + rc, you can solve the system of equations formed by equating the components of vector d on the left side with the components of the sum pa + qb + rc on the right side. Let's go step by step:
Given vector a = 2i - j + k, vector b = i - 3j + k, vector c = -2i + j - 3k, and vector d = 3i + 2j + 5k.
We want to find the values of p, q, and r such that d = pa + qb + rc.
Let's equate the components:
For the x-components: 3 = 2p + q - 2r. (Equation A)
For the y-components: 2 = -p - 3q + r. (Equation B)
For the z-components: 5 = p + q - 3r. (Equation C)
Now we have a system of equations. To solve it, we can use different methods, like substitution or elimination. Let's use the substitution method:
1. Solve Equation A for q:
q = 3 - 2p + 2r. (Equation D)
2. Substitute Equation D into Equation B:
2 = -p - 3(3 - 2p + 2r) + r.
Simplify:
2 = -p - 9 + 6p - 6r + r.
Combine like terms:
2 = 5p - 5r - 9.
Rearrange the equation:
5p - 5r = 11. (Equation E)
Now we have two equations: Equation D and Equation E. We can solve this system of equations by substitution or elimination.
Let's solve it by substitution:
3. Solve Equation E for p:
p = (11 + 5r)/5. (Equation F)
4. Substitute Equation F into Equation D:
q = 3 - 2((11 + 5r)/5) + 2r.
Simplify:
q = 3 - (22 + 10r)/5 + 2r.
Combine like terms:
q = (15 - 22 - 10r + 10r)/5.
Simplify:
q = -7/5.
Now, we have p = (11 + 5r)/5 and q = -7/5.
5. Substitute the values of p and q into any of the original equations to find the value of r.
Let's substitute it into Equation C:
5 = ((11 + 5r)/5) + (-7/5) - 3r.
Simplify:
5 = (11 + 5r - 7 - 15r)/5.
Multiply both sides by 5:
25 = 11 + 5r - 7 - 15r.
Combine like terms:
25 = -4r + 4.
Rearrange the equation:
-4r = 21.
Divide both sides by -4:
r = -21/4.
Now we have found the value of r.
6. Substitute the value of r into Equation F to find the value of p:
p = (11 + 5(-21/4))/5.
Simplify:
p = (11 - 105/4)/5.
Combine like terms:
p = (44 - 105)/20.
Simplify:
p = -61/20.
Now we have found the value of p.
7. Substitute the values of p and r into Equation D to find the value of q:
q = 3 - 2((-61/20) + 2(-21/4)) + 2(-21/4).
Simplify:
q = 3 - 2((-61/20) - 42/4) - 21/2.
Combine like terms:
q = 3 - 2((-61/20) - 42/4) - 21/2.
Simplify:
q = -53/20.
Now we have found the values of p, q, and r.
To summarize:
p = -61/20,
q = -53/20,
r = -21/4.
Therefore, the values of p, q, and r such that d = pa + qb + rc are:
p = -61/20,
q = -53/20,
r = -21/4.