# I need some help with vectors and planes.

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?

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

If the equation of plane is $$ax+by+cz=d$$ then its normal is $$$$. So, $$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)$$.
• 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 Dec 19, 2018 at 17:33