Answer :
Answer:
Vector C = 1.334i + 8.671j + 2k or 1.334x + 8.671y + 2z
Explanation:
The concept applied to solve the question is cross product of vector, AXB since vector C is perpendicular to vector A and B and this is solved by applying the 3X3 determinant method.
A detailed step by step explanation is attached below.
Answer:
[tex]C=\hat{x}1.55+\hat{y}8.54+\hat{z}2.33[/tex] and [tex]C=-\hat{x}1.55-\hat{y}8.54-\hat{z}2.33[/tex]
Explanation:
Let's write the vectors A,B and C:
[tex]A=\hat{x}2-\hat{y}+\hat{z}3[/tex]
[tex]B=\bar{x}3+\bar{y}0-\bar{z}2[/tex]
[tex]C=\hat{x}a+\hat{y}b+\hat{z}c[/tex]
Now, let's remember when two vector are perpendicular, the scalar product between them is 0. Then, using this concept, we have:
[tex]A\cdot C=(\hat{x}2-\hat{y}+\hat{z}3)(\hat{x}a+\hat{y}b+\hat{z}c)=2a-b+3c=0[/tex] (1)
[tex]B\cdot C=(\hat{x}3-\hat{y}0-\hat{z}2)(\hat{x}a+\hat{y}b+\hat{z}c)=3a-2c=0[/tex] (2)
Also we know that magnitude of C must be 9, so:
[tex]|C|^{2}=81=a^{2}+b^{2}+c^{2}[/tex] (3)
Let's solve the equation (2) for a:
[tex]a=\frac{2}{3}c[/tex] (4)
Let's put (4) on equation (1), and solve it for b:
[tex]b=\frac{2}{3}c+3c=\frac{11}{3}c[/tex] (5)
Let's put (4) and (5) in (3) and find c.
[tex]81=\left(\frac{2}{3}c\right)^{2}+\left(\frac{11}{3}c\right)^{2}+c^{2}[/tex]
[tex]c_{1}=2.33[/tex] and [tex]c_{2}=-2.33[/tex]
Finally, if we put c in (4) and (5), we will find a and b:
[tex]a_{1}=\frac{2}{3}c_{1}=1.55[/tex] and [tex]a_{2}=\frac{2}{3}c_{2}=-1.55[/tex]
[tex]b_{1}=\frac{11}{3}c_{1}=8.54[/tex] and [tex]a_{2}=\frac{11}{3}c_{2}=-8.54[/tex]
So the vector [tex]C=\hat{x}1.55+\hat{y}8.54+\hat{z}2.33[/tex] and [tex]C=-\hat{x}1.55-\hat{y}8.54-\hat{z}2.33[/tex]
I hope it helps you!