Question

If p=i+j+2k and Q=3i-2j+k,then the unit vector perpendicular to p and q is is

Answer 1

Answer:

Explanation: please find the solution in the attached photo...

Answer 2

The vector is $\bf \: \vec C= \frac{1}{ \sqrt{3} } (i + j - k )$

Explanation:

Given:

• Two vectors $\vec P = i + j + 2k$
• and $\vec Q= 3i - 2j + k$

To find:

• Find the unit vector perpendicular to P and Q.

Solution:

Concept to be used:

A unit vector C which is perpendicular to P and Q is given by

$\bf \vec C = \frac{\vec P \times\vec Q }{ |\vec P \times\vec Q| }$

here

$\vec P \times\vec Q$ is cross products of vectors P and Q.

and $|\vec P \times\vec Q|$ is magnitude of cross product.

Step 1:

Find the cross product of vector P and Q.

$\vec P \times\vec Q = \left|\begin{array}{ccc}i&j&k\\1&1&2\\3&-2&1\end{array}\right|$

Simplify the determinant of right hand side,

$= i(1 + 4) - j(1 - 6) + k( - 2 - 3)$

or

$\bf\vec P \times\vec Q= 5i + 5j - 5k$

Step 2:

Find the magnitude of cross product.

$|\vec P \times\vec Q| = \sqrt{( {5}^{2}) + ( {5)}^{2} + ( { - 5)}^{2} }$

or

$|\vec P \times\vec Q| = \sqrt{25 + 25 + 25}$

or

$|\vec P \times\vec Q| = \sqrt{75}$

or

$\bf\:|\vec P \times\vec Q| = 5 \sqrt{3}$

Step 3:

Find the unit vector perpendicular to P and Q.

$\vec C = \frac{5i + 5j - 5k}{5 \sqrt{3} }$

or

$\vec C = \frac{i + j - k}{ \sqrt{3} }$

or

$\bf \: \vec C= \frac{1}{ \sqrt{3} } (i + j - k )$

Thus,

The vector is $\bf \: \vec C = \frac{1}{ \sqrt{3} } (i + j - k )$

Learn more:

1) Let a = 2i + 4j -5k, b = i + j + k and c = j +2k. Find the unit vector in the opposite direction of a + b + c.

https://brainly.in/question/7029634

2) Find a vector of magnitude 3 and perpendicular to both the vectors b = 2i - 2j + k and c = 2i + 2j + 3k.

https://brainly.in/question/7029678

#SPJ3