Some vector algebra problems.

**GreenMile** · April 14th 2009, 10:03 PM

1.Find $\vec{a} x \vec{b}$ (Vectorial product) if $\vec{a}=3\vec{m}+2\vec{n}$ , $\vec{b}=\vec{m}+5\vec{n}$ and $|\vec{m}|=|\vec{n}|=1$ . (Angle between m and n is 90 degrees.)

2.We have vectors $\vec{a}=(0,2\lambda,\lambda),$ $\vec{b}=(2,2,1)$ and $\vec{c}=(-1,-2,-1)$

a) Find $\vec{d}$ from this conditions: $\vec{a}x\vec{b} = \vec{c}x\vec{d}$ and $\vec{a}x\vec{c} = \vec{b}x\vec{d}$ (Vectorial product).
b) Prove that vectors $\vec{a} - \vec{d}$ and vectors $\vec{b} - \vec{c}$ are colinear.
c) Find $\lambda$ from condition $(\vec{a}-\vec{b})*\vec{c}=\vec{a}*\vec{c}+\lambda$

3. Write equation of normal from point $M(2,3,1)$ into line: $l: \frac{x+1}{2}=\frac{y-0}{-1}=\frac{z-2}{3}$ .

Thank you.

**Logic** · April 15th 2009, 02:53 AM

The first one should go like shown below.

**GreenMile** · April 15th 2009, 01:44 PM

logic, it's vectorial (cross) product. Any idea for these problems?

Anyway thank you.

**mr fantastic** · April 28th 2009, 10:13 PM

Originally Posted by GreenMile

1.Find $\vec{a} x \vec{b}$ (Vectorial product) if $\vec{a}=3\vec{m}+2\vec{n}$ , $\vec{b}=\vec{m}+5\vec{n}$ and $|\vec{m}|=|\vec{n}|=1$ . (Angle between m and n is 90 degrees.)

[snip]

Note that m x m = n x n = 0. Nothing much can be said about m x n except that it will be another vector perpendicular to m and n and that its magnitude will be equal to 1.

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