Question

Given vectors A = xˆ2−yˆ +zˆ3 and B = xˆ3−zˆ2, find a vector C whose magnitude is 9 and whose direction is perpendicular to both A and B

Answer 1

Answer:

$C=\hat{x}1.55+\hat{y}8.54+\hat{z}2.33$ and $C=-\hat{x}1.55-\hat{y}8.54-\hat{z}2.33$

Explanation:

Let's write the vectors A,B and C:

$A=\hat{x}2-\hat{y}+\hat{z}3$

$B=\bar{x}3+\bar{y}0-\bar{z}2$

$C=\hat{x}a+\hat{y}b+\hat{z}c$

Now, let's remember when two vector are perpendicular, the scalar product between them is 0. Then, using this concept, we have:

$A\cdot C=(\hat{x}2-\hat{y}+\hat{z}3)(\hat{x}a+\hat{y}b+\hat{z}c)=2a-b+3c=0$ (1)

$B\cdot C=(\hat{x}3-\hat{y}0-\hat{z}2)(\hat{x}a+\hat{y}b+\hat{z}c)=3a-2c=0$ (2)

Also we know that magnitude of C must be 9, so:

$|C|^{2}=81=a^{2}+b^{2}+c^{2}$ (3)

Let's solve the equation (2) for a:

$a=\frac{2}{3}c$ (4)

Let's put (4) on equation (1), and solve it for b:

$b=\frac{2}{3}c+3c=\frac{11}{3}c$ (5)

Let's put (4) and (5) in (3) and find c.  

$81=\left(\frac{2}{3}c\right)^{2}+\left(\frac{11}{3}c\right)^{2}+c^{2}$

$c_{1}=2.33$ and $c_{2}=-2.33$  

Finally, if we put c in (4) and (5), we will find a and b:

$a_{1}=\frac{2}{3}c_{1}=1.55$ and $a_{2}=\frac{2}{3}c_{2}=-1.55$  

$b_{1}=\frac{11}{3}c_{1}=8.54$ and $a_{2}=\frac{11}{3}c_{2}=-8.54$

So the vector $C=\hat{x}1.55+\hat{y}8.54+\hat{z}2.33$ and $C=-\hat{x}1.55-\hat{y}8.54-\hat{z}2.33$

I hope it helps you!

Answer 2

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.