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Unit Vectors

Find the unit...

Question

Find the unit vectors along $\to A$ and $\to B$ where $\to A = 3^i - 2^j + 4^k$ and $\to B = 4^i - 2^j + 3^k$

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Solution

Unit Vector,
$\begin{matrix} ^A = \frac{\to A}{| A |} = \frac{3^i - 2^j + 4^k}{\sqrt{26}}^B = \frac{\to B}{| B |} = \frac{4^i - 2^j + 3^k}{\sqrt{29}} \end{matrix}$

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Similar questions

\to a

\to b

are vectors such that

| \to a + \to b | = \sqrt{29}

\to a \times (2^i + 3^j + 4^k) = (2^i + 3^j + 4^k) \times \to b

then a possible value of

(\to a + \to b) \cdot (- 7^i + 2^j + 3^k)

\to a

\to b

are vectors such that

| \to a + \to b | = \sqrt{29}

\to a \times (2^i + 3^j + 4^k) = (2^i + 3^j + 4^k) \times \to b

, then a possible value of

(\to a + \to b) . (- 7^i + 2^j + 3^k)

Q. Find the cosine of the angle between the directions of the vectors

\to a = 4^i + 3^j +^k, \to b = 2^i -^j + 2^k .

Also find a unit vector perpendicular to both

\to a

\to b

. What is the sine of the angle between

\to a

\to b

\to a =^i + 4^j + 2^k, \to b = 3^i - 2^j + 7^k

\to c = 2^i -^j + 4^k

. Find a vector

\to d

which is perpendicular to both

\to a

\to b

\to c \cdot \to d = 15

\to A = 2^i + 3^j + 6^k

\to B = 3^i - 6^j + 2^k

then vector perpendicular to both

\to A

\to B

K

6^i + 2^j - 3^k

K =

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