Question

Let $L_1: x = y = z$ and $L_2: x-1=y-2=z-3$ be two lines. The foot of perpendicular drawn from the origin $O(0,0,0)$ on $L_2$ is $A$. If the equation of a plane containing the line $L_1$ and perpendicular to $OA$ is $10x + by + cz = d$, then the value of $b + c + d$ is equal to

Answer 1

-10

Answer 2

The given lines are $L_1: x = y = z$ and $L_2: x-1=y-2=z-3$.

$L_1$ can be written as $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$. It passes through the origin $O(0,0,0)$ and has direction vector $\vec{v}_1 = (1,1,1)$.

$L_2$ can be written as $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{1}$. It passes through the point $P_2(1,2,3)$ and has direction vector $\vec{v}_2 = (1,1,1)$.

Note that $L_1$ and $L_2$ are parallel lines since their direction vectors are the same.

We need to find the foot of the perpendicular $A$ from the origin $O(0,0,0)$ to the line $L_2$. Any point on $L_2$ can be represented as $A(1+s, 2+s, 3+s)$ for some parameter $s$. The vector $\vec{OA}$ from $O(0,0,0)$ to $A(1+s, 2+s, 3+s)$ is $\vec{OA} = (1+s, 2+s, 3+s)$. Since $A$ is the foot of the perpendicular from $O$ to $L_2$, the vector $\vec{OA}$ must be perpendicular to the direction vector of $L_2$, $\vec{v}_2 = (1,1,1)$. The dot product $\vec{OA} \cdot \vec{v}_2$ must be zero: $(1+s)(1) + (2+s)(1) + (3+s)(1) = 0$ $1+s + 2+s + 3+s = 0$ $6 + 3s = 0$ $3s = -6$ $s = -2$.

Substitute $s=-2$ into the coordinates of $A$: $A(1+(-2), 2+(-2), 3+(-2)) = A(-1, 0, 1)$. So, the foot of the perpendicular from the origin to $L_2$ is $A(-1, 0, 1)$.

The vector $\vec{OA}$ is $\vec{OA} = A - O = (-1, 0, 1) - (0,0,0) = (-1, 0, 1)$.

We need to find the equation of a plane that contains the line $L_1$ and is perpendicular to $\vec{OA}$. The plane contains the line $L_1$. This means:

• The plane passes through any point on $L_1$. Since $L_1$ passes through the origin $O(0,0,0)$, the plane also passes through $O(0,0,0)$.
• The direction vector of $L_1$, $\vec{v}_1 = (1,1,1)$, is parallel to the plane.

The plane is perpendicular to the vector $\vec{OA} = (-1, 0, 1)$. This means $\vec{OA}$ is a normal vector to the plane. Let the normal vector of the plane be $\vec{n}$. We can choose $\vec{n}$ to be proportional to $\vec{OA}$. Let $\vec{n} = k \vec{OA} = k(-1, 0, 1) = (-k, 0, k)$ for some non-zero scalar $k$.

The equation of a plane passing through a point $P_0(x_0, y_0, z_0)$ with normal vector $\vec{n}=(A,B,C)$ is $A(x-x_0) + B(y-y_0) + C(z-z_0) = 0$. Here, the plane passes through $O(0,0,0)$ and has normal vector $\vec{n}=(-k, 0, k)$. So the equation of the plane is: $-k(x-0) + 0(y-0) + k(z-0) = 0$ $-kx + kz = 0$. Assuming $k \neq 0$, we can divide by $-k$: $x - z = 0$.

We are given that the equation of the plane is $10x + by + cz = d$. The equations $x - z = 0$ and $10x + by + cz = d$ represent the same plane. We can multiply the first equation by 10 to match the coefficient of $x$: $10(x - z) = 10(0)$ $10x - 10z = 0$.

Comparing this with $10x + by + cz = d$: The coefficient of $x$ is 10 in both equations. The coefficient of $y$ in $10x - 10z = 0$ is 0. So, $b=0$. The coefficient of $z$ in $10x - 10z = 0$ is -10. So, $c=-10$. The constant term in $10x - 10z = 0$ is 0. So, $d=0$.

We need to find the value of $b+c+d$. $b+c+d = 0 + (-10) + 0 = -10$.