Question #52240

IF line(OA) =4i + 3j, line(OB) = 6i - 2j and line(OC) =2i - j . Deduce the length of the triangle CA.

13−−√

20−−√

29−−√

17−−√
2 What quantity have both magnitude and direction ?
vector
scalar
mass
dot product
3 Find the vector product axb. If a = i + 2j - k and b = 2i + 3j + k
5i - 3j - k
2i - 4j - k
3i + j - k
I - j + 3k
4 a and b are vectors defined by a = 8i + 2j - 3k and b = 3i - 6j + 4k, where I,j,k are mutually perpendicular unit vectors. Show that a and b are perpendicular to each other.
90
45
1
0
5 a and b are vectors defined by a = 8i + 2j - 3k and b = 3i - 6j + 4k, where I,j,k are mutually perpendicular unit vectors. Calculate a and b.
1
0
2
4

Expert's answer

Answer on Question #52240 – Math – Vector Calculus

1. IF line(OA) = 4i + 3j, line(OB) = 6i - 2j and line(OC) = 2i - j. Deduce the length of the triangle CA.

13--v

20--v

29--v

17--v

2. What quantity have both magnitude and direction?

vector

scalar

mass

dot product

3. Find the vector product axb. If a = i + 2j - k and b = 2i + 3j + k

5i - 3j - k

2i - 4j - k

3i + j - k

1 - j + 3k

4. A and b are vectors defined by a = 8i + 2j - 3k and b = 3i - 6j + 4k, where 1, j, k are mutually perpendicular unit vectors. Show that a and b are perpendicular to each other.

90

45

1


5 a and b are vectors defined by a=8i+2j3ka = 8i + 2j - 3k and b=3i6j+4kb = 3i - 6j + 4k, where i,j,ki, j, k are mutually perpendicular unit vectors. Calculate a and b.

1


2

4

Solution

1. AC=OCOA=2ij(4i+3j)=2i4j\overline{AC} = \overline{OC} - \overline{OA} = 2i - j - (4i + 3j) = -2i - 4j

AC=22+42=20|\overline{AC}| = \sqrt{2^2 + 4^2} = \sqrt{20}


2. vector

3. Cross (or vector) product


a×b=ijk121231=2131i1121j+1223k==5i3jk\begin{array}{l} a \times b = \left| \begin{array}{ccc} i & j & k \\ 1 & 2 & -1 \\ 2 & 3 & 1 \end{array} \right| = \left| \begin{array}{cc} 2 & -1 \\ 3 & 1 \end{array} \right| i - \left| \begin{array}{cc} 1 & -1 \\ 2 & 1 \end{array} \right| j + \left| \begin{array}{cc} 1 & 2 \\ 2 & 3 \end{array} \right| k = \\ = 5i - 3j - k \end{array}


4. Dot (or scalar) product


ab=83+2(6)+(3)4=241212=0ab = 8 * 3 + 2 * (-6) + (-3) * 4 = 24 - 12 - 12 = 0


Angle between a and b equals 9090{}^\circ.

5. Dot (or scalar) product


ab=83+2(6)+(3)4=241212=0ab = 8 * 3 + 2 * (-6) + (-3) * 4 = 24 - 12 - 12 = 0


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