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My question says:

Consider the vectors $a = i - j + k, b = i + 2j + 4k$ and $c = 2i - 5j - k.$

a) Given that $c = ma + nb$ where $m, n$ are real numbers, find the value of $m$ and $n$

b) Find a unit vector, $u$, normal to both $a$ and $b.$

For $(a)$ I have done this and got $n = -1$ and $m = 3$.

For $(b)$ I got $\frac{1}{\sqrt{118}}\left(-10i-3j+3k\right)$ as my unit vector.

But now I’m faced with

c) The plane $p_1$ contains the point $A (1, -1, 1)$ and is normal to $b.$ The plane intersects the x, y and z axes at the points $L, M$ and $N$ respectively:

i) Find the Cartesian equation of $p_1$

ii) Write doen the coordinates of $L, M$ and $N$

I’m not sure where to start at this point, can anyone help?

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  • $\begingroup$ But $$\vec{c}$$ is given, or i'm missing something? $\endgroup$ Dec 19, 2018 at 16:22
  • $\begingroup$ Only vector c is given $\endgroup$
    – D. Wetzel
    Dec 19, 2018 at 16:51

1 Answer 1

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If the equation of plane is $ax+by+cz=d$ then its normal is $<a,b,c>$. So, $p_1$'s equation is $$1(x-1)+2(y+1)+4(z-1)=0$$ or $$x+2y+4z=3$$ which contains the point $A$. Converting it to intercept form: $$\frac x3+\frac y{\frac32}+\frac z{\frac34}=1$$ which gives $L=(3,0,0);M=\Bigl(0,\frac32,0\Bigr);N=\Bigl(0,0,\frac34\Bigr)$.

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  • $\begingroup$ Could you please explain how you derived the first equation? $\endgroup$
    – D. Wetzel
    Dec 19, 2018 at 17:27
  • $\begingroup$ Taking the dot product of a vector on the plane and its normal gives you $0$. So assume the normal to be standing over the point $A$ and also an arbitrary point $(x,y,z)$ on the plane. Now the vector on the plane is simply $<x-1,y+1,z-1>$. Take its dot product with the normal to get a relation between the values of $x,y,z$ which is nothing but the equation of plane $\endgroup$ Dec 19, 2018 at 17:33

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