Let $L$ be the line of intersection of the planes $ax+by+cz=1$ and $a'x+b'y+c'z=1$.Find the perpendicular distance of $L$ from the origin.

Question

Let $L$ be the line of intersection of the planes $ax+by+cz=1$ and $a'x+b'y+c'z=1$.

Also,given that $$(a-a')^2+(b-b')^2+(c-c')^2=16$$ and $$(ab'-a'b)^2+(bc'-b'c)^2+(ac'-a'c)^2=9$$

Find the perpendicular distance of $L$ from the origin.

My Attempt

Let the direction ratios of the line $L$ be $l,m,n$ where

$al+bm+cn=0$

and $a'l+b'm+c'n=0$.

Solving for $l,m,n$

$l=bc'-b'c$

$m=a'c-ac'$

$n=ab'-a'b$

Let $P(\alpha,\beta,\gamma)$ be the foot of perpendicular dropped upon $L$ from the origin.

$a\alpha+b\beta+c \gamma=1$

$a'\alpha+b'\beta+c'\gamma=1$

which leads to $(a-a')\alpha+(b-b')\beta+(c-c')\gamma=0$

But beyond this I am not able to solve. How is one supposed to link all this information.

Answer 1

Let us convert the question in vector form. It can be read as follows:

Let $L$ be line of intersection of planes $\vec r.\vec a=1$ and $\vec r.\vec b=1$, where $|\vec a-\vec b|=4$ and $|\vec a \times \vec b|=3$. Find perpendicular distance of the line $L$ from the origin.

Let $\vec c$ be the position vector of the foot of perpendicular dropped upon the line from origin. So we have to find $|\vec c|.$

Since the line is perpendicular to both $\vec a$ and $\vec b$,therefore the line must be parallel to $\vec a\times\vec b.$

Therefore,$\vec c$ is perpendicular to $\vec a\times\vec b$.

Also $\vec c.\vec a=1$ and $\vec c.\vec b=1$ since $\vec c$ lies on $L$ which in turn lies on both the planes

So,

$\vec c.(\vec a\times\vec b)=0$

$[\vec c\space \vec a \space \vec b]=0$

Thus $\vec c$ , $\vec a$ and $\vec b$ are coplanar and thus $\vec c$ can be expressed as a linear combination of $\vec a$ and $\vec b$.

$\vec c=\lambda \vec a+\mu \vec b$

Taking scalar(dot) product with $\vec c$ on both sides, we get

$\vec c.\vec c=\lambda \vec a.\vec c+\mu \vec b.\vec c$

$|\vec c|^2=\lambda+\mu$

Similarly, taking dot product with $\vec a$ and then by $\vec b$ on both sides we obtain the two equations

$\lambda a^2+\mu \vec a.\vec b=1$

$\lambda \vec a.\vec b+\mu b^2=1$

Solving for $\lambda$ and $\mu$ from the above two equations

$$\lambda=\frac{b^2-\vec a.\vec b}{a^2b^2-(\vec a.\vec b)^2}=\frac{b^2-\vec a.\vec b}{|\vec a\times\vec b|^2}$$

$$\mu=\frac{a^2-\vec a.\vec b}{a^2b^2-(\vec a.\vec b)^2}=\frac{a^2-\vec a.\vec b}{|\vec a\times\vec b|^2}$$

$$|\vec c|^2=\lambda+\mu=\frac{b^2-\vec a.\vec b}{|\vec a\times\vec b|^2}+\frac{a^2-\vec a.\vec b}{|\vec a\times\vec b|^2}=\frac{|\vec a-\vec b|^2}{|\vec a\times\vec b|^2}=\frac{16}{9}$$

So,$|\vec c|=\frac{4}{3}$

Answer 2

Think of it as a constrained optimization problem.

$$\min x^2+y^2+z^2$$ s.t. $$ax+by+cz=1$$ $$a'x+b'y+c'z=1$$

The Lagrangian becomes:

$$L = x^2+y^2+z^2-\lambda(ax+by+cz-1)-\mu(a'x+b'y+c'z-1)$$

Taking derivative with respect to $x$ and setting to $0$,

$$\hat{x}=\frac{a\lambda+a'\mu}{2}$$

and similarly for $\hat{y}$ and $\hat{z}$.

where the optimal point is $(\hat{x}, \hat{y}, \hat{z})$

Note that this is the same as the condition $\vec{c}=\lambda \vec{a}+\mu \vec{b}$ in the answer by @Maverick.

Now, the constraints must be satisfied, so

$$a\hat{x}+b\hat{y}+c\hat{z}=1$$ and $$a'\hat{x}+b'\hat{y}+c'\hat{z}=1$$ Use these to get $\lambda$ and $\mu$.