Suppose A=3i-2j+k, B=-i-4j+3k&C is a vector lying along x-axis with the property that A.(B+C)=0. Then find vector C
Well, if A.(B+C) equals zero, it means that the dot product of vector A and the sum of vectors B and C is zero. Let's calculate it step by step:
A = 3i - 2j + k
B = -i - 4j + 3k
C = ?
First, let's find B + C:
(B + C) = (-i - 4j + 3k) + (C1i + C2j + C3k)
To make the dot product zero, we know that the coefficients of i, j, and k must add up to zero. Therefore:
-1 + C1 = 0 (for i)
-4 + C2 = 0 (for j)
3 + C3 = 0 (for k)
Simplifying these equations, we find:
C1 = 1
C2 = 4
C3 = -3
Hence, vector C can be written as C = 1i + 4j - 3k.
To find vector C, we first need to calculate the dot product of A with B+C, and then solve for C.
Given:
A = 3i - 2j + k,
B = -i - 4j + 3k.
Let's assume C = xi + yj + zk, where x, y, and z are variables to be determined.
Now, we can find the dot product of A with B+C:
A.(B+C) = 0.
Substituting the values of A, B, and C into the equation, we get:
(3i - 2j + k).((-i - 4j + 3k) + (xi + yj + zk)) = 0.
Expanding the equation:
(3i - 2j + k).(-i - 4j + 3k + xi + yj + zk) = 0.
Now, distribute the dot product:
(3 * -1) + (3 * -4) + (3 * 3) + (3 * x) + (-2 * -1) + (-2 * -4) + (-2 * 3) + (-2 * y) + (1 * -1) + (1 * -4) + (1 * 3) + (1 * z) = 0.
Simplifying further:
-3 - 12 + 9 + 3x + 2 + 8 - 6 - 2y - 1 - 4 + 3 + z = 0.
Combining like terms:
-3x - 2y + z - 5 = 0.
Therefore, we have the system of equations:
-3x - 2y + z = 5.
Since the vector C lies along the x-axis, the y and z components are both equal to 0. So, y = 0 and z = 0.
Now, we can solve the system of equations to find x:
-3x - 2(0) + 0 = 5.
-3x = 5.
Divide both sides by -3:
x = -5/3.
Therefore, the vector C is:
C = (-5/3)i + 0j + 0k.
Hence, the vector C is given by C = (-5/3)i.
To find vector C, we need to determine the value of C such that A · (B + C) = 0.
First, let's substitute the given values for A and B into the equation:
A · (B + C) = 0
(3i - 2j + k) · (-i - 4j + 3k + C) = 0
Next, we can expand and simplify the dot product:
A · (B + C) = (3i) · (-i - 4j + 3k + C) + (-2j) · (-i - 4j + 3k + C) + (k) · (-i - 4j + 3k + C) = 0
Simplifying further:
(-3i - 12j + 9k - 3Ci) + (2j + 8i - 6k - 2Cj) + (k + 4j - 3i - Ck) = 0
Combining like terms:
(-3i + 8i - 3i - 3Ci) + (2j - 12j + 4j - 2Cj) + (k - 6k + 9k - Ck) = 0
Simplifying:
(2i + 2j + 5k - 3Ci - 2Cj - Ck) = 0
Now, since the dot product A · (B + C) = 0, the sum of the coefficients of i, j, k, -Ci, -Cj, and -Ck must equal zero.
So we have the following equations:
2 - 3C = 0 (coefficient of i)
2 - 2C = 0 (coefficient of j)
5 - C = 0 (coefficient of k)
Solving these equations:
2 - 3C = 0
=> 3C = 2
=> C = 2/3
2 - 2C = 0
=> 2C = 2
=> C = 1
5 - C = 0
=> C = 5
From these equations, we find that C = 2/3i + j + 5k.
Therefore, the vector C = 2/3i + j + 5k.
B+C = (x-1)i-4j+3k
A•(B+C) = 3(x-1)i
so x = 1 and C = 1i