Question

What are vectors that are not parrallel to the same line,called.
scalar
collinear vectors
non-collinear vectors
vectors
7 Find the vector product axb. If a = 2i + 3j + 4k and b = 5i - 2j + k
11i + 18j - 19k
2j + 3k
5i - 6j + 7k
4i - 6j +11k
8 Find the scalar product a.b. If a = 2i + 3j + 4k and b = 5i - 2j + k
9
10
8
7
9 The centroid of a triangle of the triangle OAB is denoted by G. If o is the origin and line(OA) = 4i + 3j, line(OB) = 6i - j,find line(OG) in terms of the unit vectors I and j
10i - 3j

12(10i−2j)
10i + 2j

13(10i+2j)
10 Given that a = 5i +2j - k and b = I - 3j + k. Find (a + b) x (a + b).
2i - 12j - 34k
2i + 12j + 34k
2i -3j + 12j
2i + 2k

Answer 1

Answer on Question #52241 – Math – Vector Calculus

What are vectors that are not parallel to the same line, called?

- scalar

- collinear vectors

- non-collinear vectors

- vectors

Answer: non-collinear vectors.

7 Find the vector product axb. If $a = 2i + 3j + 4k$ and $b = 5i - 2j + k$

$$11i + 18j - 19k$$ $$2j + 3k$$ $$5i - 6j + 7k$$ $$4i - 6j + 11k$$

Solution

$$\vec{a}x\vec{b} = \begin{vmatrix} i & j & k \\ 2 & 3 & 4 \\ 5 & -2 & 1 \end{vmatrix} = i(3 \cdot 1 - 4(-2)) + j(4 \cdot 5 - 1(2)) + k(2 \cdot (-2) - 3(5)) = 11i + 18j - 19k.$$

Answer: $11i + 18j - 19k$.

8 Find the scalar product a.b. If $a = 2i + 3j + 4k$ and $b = 5i - 2j + k$

9

10

8

7

Solution

$$\vec{a} \cdot \vec{b} = (2i + 3j + 4k)(5i - 2j + k) = 2 \cdot 5 + 3 \cdot (-2) + 4 \cdot 1 = 8.$$

Answer: 8.

9 The centroid of a triangle of the triangle OAB is denoted by G. If o is the origin and line(OA) = $4i + 3j$, line(OB) = $6i - j$, find line(OG) in terms of the unit vectors I and j

$$10i - 3j$$

1/2(10i-2j)

10i + 2j

1/3(10i+2j)

Solution

Vector $\overline{OB} = \overline{OA} + \overline{AB}$, hence $\overline{AB} = \overline{OB} - \overline{OA}$, $\overline{AM} = \frac{1}{2}\overline{AB}$.

Vector $\overline{OM} = \overline{OA} + \overline{AM} = \overline{OA} + \frac{1}{2}\overline{AB} = \overline{OA} + \frac{1}{2}(\overline{OB} - \overline{OA}) = \frac{1}{2}(\overline{OB} + \overline{OA})$.

Let OM be the median of the triangle OAB. By properties of centroid, $OG = \frac{2}{3} OM$.

Thus,

$$\overline{OG} = \frac{2}{3} \overline{OM} = \frac{2}{3} \cdot \frac{1}{2} (\overline{OB} + \overline{OA}) = \frac{1}{3} (4i + 3j + 6i - j) = \frac{1}{3} (10i + 2j).$$

Answer: 1/3(10i+2j).

10 Given that $a = 5i + 2j - k$ and $b = 1 - 3j + k$. Find $(a + b) \times (a - b)$.

2i - 12j - 34k

2i + 12j + 34k

2i - 3j + 12j

2i + 2k

Solution

$$(\overline{a + b}) = 5i + 2j - k + i - 3j + k = 6i - j.$$ $$(\overline{a - b}) = 5i + 2j - k - i + 3j - k = 4i + 5j - 2k.$$ $$(\overline{a + b}) \times (\overline{a - b}) = \begin{vmatrix} i & j & k \\ 6 & -1 & 0 \\ 4 & 5 & -2 \end{vmatrix} = 2i + 12j + 34k.$$

Answer: $2i + 12j + 34k$.

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