Why we can't remove the parameter and find the Cartesian equation of straight lines in higher dimensions ?
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1$\begingroup$ I don't really understand what this book means, but certainly you can represent a line in $n$-D space by $n-1$ linear equations. No parameter $t$ is involved. $\endgroup$– Troy WooCommented Oct 12, 2014 at 21:12
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$\begingroup$ the book is introduction to linear algebra by serge lang $\endgroup$– Mohamed OsamaCommented Oct 12, 2014 at 21:14
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1 Answer
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If we only have the two equations $$(1)\ x_1=p_1+ta_1$$ $$(2)\ x_2=p_2+ta_2$$
we can eliminate $t$ and get the equation $$(x_1-p_1)a_2-(x_2-p_2)a_1=0$$ which we can transform into $$ax_1+bx_2=c$$ This equation is true, no matter which value $t$ has.
If we add the equation $$(3)\ x_3=p_3+ta_3$$
the value of $x_3$ depends on the value $t$, so an equation of the form $$ax_1+bx_2+cx_3=d$$
cannot hold for arbitary $t$ because if $x_1$ is given, $x_2$ is fixed.
Alternatively, you can think about the meaning of the normal vector. It is a vector which is orthognal to the line. The direction of this vector must be unique, which is the case in $2$ dimensions, but not in $3$ or more. There are infinitely many vectors with different directions which are orthogonal to a given line in $3$ or more dimensions.
