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Vectors and Its Types

The magnitude...

Question

The magnitude of a given vector with end points (4, -4, 0) and (-2, -2, 0) must be:

6

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5 \sqrt{2}

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4

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2 \sqrt{10}

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Solution

The correct option is D $2 \sqrt{10}$
$\to r = {\to r}_{2} - {\to r}_{1} = (- 2^i - 2^j + 0^k) - (4^i - 4^j + 0^k)$
$\Rightarrow \to r = - 6^i + 2^j + 0^k$
$∴ | \to r | = \sqrt{(- 6)^{2} + (2)^{2} + 0^{2}} = \sqrt{36 + 4} = \sqrt{40} = 2 \sqrt{10}$ .

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The magnitude of a given vector with end points (4, -4, 0) and (-2, -2, 0) must be

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 4 6 & 2 & 8 2 & 4 & 6 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 8 & 4 & - 2 0 & - 2 & 0 2 & 2 & 6 \end{matrix} ⎤ ⎥ ⎦

C = ⎡ ⎢ ⎣ \begin{matrix} 8 & 2 & 0 0 & - 2 & - 6 - 8 & 4 & - 10 \end{matrix} ⎤ ⎥ ⎦

A - B - C

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 4 6 & 2 & 8 2 & 4 & 6 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 8 & 4 & - 2 0 & - 2 & 0 2 & 2 & 6 \end{matrix} ⎤ ⎥ ⎦

C = ⎡ ⎢ ⎣ \begin{matrix} 8 & 2 & 0 0 & - 2 & - 6 - 8 & 4 & - 10 \end{matrix} ⎤ ⎥ ⎦

A + B + C

A = ⎡ ⎢ ⎣ \begin{matrix} 2 & 0 & 4 6 & 2 & 8 2 & 4 & 6 \end{matrix} ⎤ ⎥ ⎦, B = ⎡ ⎢ ⎣ \begin{matrix} 8 & 4 & - 2 0 & - 2 & 0 2 & 2 & 6 \end{matrix} ⎤ ⎥ ⎦

C = ⎡ ⎢ ⎣ \begin{matrix} 8 & 2 & 0 0 & - 2 & - 6 - 8 & 4 & - 10 \end{matrix} ⎤ ⎥ ⎦

(3 A + 2 B) - 3 C

- 2, 2, 6, 10, . . .

\frac{2}{3}

a = 18, n = 10, a_{n} = 0

- 5

a = 0, a_{10} = 6

4

a_{2} = 13, a_{4} = 3

- 4

- 2

\frac{1}{2}

5

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