Suppose vector A=3i_2j+k ; vector B=-i-4j+3k and vector C is lyingbalong x-axis with the property that A.(B+C)=0,then calculate the vector C.
C = x i + 0 j + 0 k
I assume you mean A = 3 i - 2 j + 1 k
and B = -1 i - 4 j + 3 k
then B + C = (x-1) i +( 0-4) j + (0+3) k
Now if you mean A dot (B+C) =0
3(x-1) -2 ( -4) + (3) = 0
3 x -3 + 8 + 3 = 0
x = -8/3
so C = -8/3 i + 0j + 0 k
check my arithmetic
To find vector C, we need to use the given property that A · (B + C) = 0.
First, let's calculate what A · (B + C) is:
A · (B + C) = A · B + A · C
A · B = (3i + 2j + k) · (-i - 4j + 3k)
= -3i - 8j + 3k
A · (B + C) = (-3i - 8j + 3k) + A · C
= -3i - 8j + 3k + A · C
According to the given property, A · (B + C) equals 0. So, we can set the equation equal to zero:
-3i - 8j + 3k + A · C = 0
Now, let's express vector A in terms of its components:
A = 3i + 2j + k
Substituting this into the equation:
-3i - 8j + 3k + (3i + 2j + k) · C = 0
Expanding the dot product:
-3i - 8j + 3k + (3i + 2j + k) · C = 0
-3i - 8j + 3k + (3iC + 2jC + kC) = 0
-3i - 8j + 3k + (3C)i + (2C)j + (C)k = 0
Comparing the coefficients of i, j, and k on both sides:
-3 + 3C = 0 (coefficient of i)
-8 + 2C = 0 (coefficient of j)
3 + C = 0 (coefficient of k)
Solving these equations, we get:
C = -3 (from the first equation)
C = 4 (from the second equation)
C = -3 (from the third equation)
Therefore, vector C = -3i + 4j - 3k.
To find the vector C that satisfies the condition A.(B+C) = 0, we can start by expanding the dot product expression:
A.(B + C) = 0
Using the distributive property of dot product over addition, we get:
A.B + A.C = 0
Next, let's calculate the dot product A.B:
A.B = (3i + 2j + k).(-i - 4j + 3k)
= 3(-1) + 2(-4) + 1(3)
= -3 - 8 + 3
= -8
Now, substitute the value of A.B into the previous equation:
-8 + A.C = 0
To isolate A.C, we move -8 to the other side:
A.C = 8
Since vector C is stated to lie along the x-axis, it has no y or z-components. Therefore, the vector C can be expressed as:
C = Cx i
Now, substitute C into the equation A.C = 8:
(3i + 2j + k) . (Cx i) = 8
This can be further simplified by distributing the dot product:
3Cx + 2(0) + 1(0) = 8
3Cx = 8
Solve for Cx:
Cx = 8/3
Therefore, the vector C can be written as:
C = (8/3)i