Q.Consider the vectors $\vec a,\vec b,\vec c,\vec d$ such that $(\vec a\times\vec b)\times(\vec c\times\vec d)=\vec 0$. Let $P_1$ and $P_2$ be the planes determined by the pairs of vectors $\vec a,\vec b$ and $\vec c,\vec d$ respectively. Then the angle between $P_1$ and $P_2$ is

The foot of the perpendicular from a point to a plane is exactly the midpoint of that point and its reflected image in the plane.

1. When a point $A$ is reflected in a plane to get its image $A'$, the line segment $AA'$ is perpendicular to the plane, and the plane bisects $AA'$.
2. So the foot of the perpendicular from $A$ to the plane is the midpoint $M$ of $A$ and $A'$. …

Extracting each line's direction vector (rewriting the second line's $2y+3$ term into standard symmetric form) and finding their dot product is zero shows the lines are perpendicular.

1. Line 1: $\dfrac{x-2}{3}=\dfrac{y+1}{-2}$, $z=2$. Since $z$ is fixed (no variation with the parameter), its direction ratios are $\vec d_1=(3,-2,0)$.
2. Line 2: $\dfrac{x-1}{1}=\dfrac{2y+3}{3}=\dfrac{z+5}{2}$. Rewrite the middle term: $2y+3=2\left(y+\dfrac32\right)$, so $\dfrac{2y+3}{3}=\dfrac{y+3/2}{3/2}$. …

The angle between a line and a plane uses $\sin\theta=|\vec d\cdot\vec n|/(|\vec d||\vec n|)$ with the line's direction and the plane's normal.

1. Line direction: $\vec d=(2,1,-2)$ (from $\vec r=(\hat i+2\hat j-3\hat k)+t(2\hat i+\hat j-2\hat k)$).
2. Plane $\vec r\cdot(\hat i+\hat j)+4=0$ has normal $\vec n=(1,1,0)$.
3. $\vec d\cdot\vec n=2(1)+1(1)+(-2)(0)=3$. $|\vec d|=\sqrt{4+1+4}=3$. $|\vec n|=\sqrt{1+1}=\sqrt2$. …

Using the point-to-plane distance formula on $3x-6y+2z+7=0$ from the origin.

1. For a plane $ax+by+cz+d=0$, the perpendicular distance from a point $(x_0,y_0,z_0)$ is $\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$.
2. Here $a=3,\,b=-6,\,c=2,\,d=7$ and the point is the origin $(0,0,0)$. …

The direction-ratio dot product of the two lines is zero, so the lines are perpendicular and the angle between them is $\dfrac{\pi}{2}$.

1. The first line $\dfrac{x-4}{2}=\dfrac{y}{1}=\dfrac{z+1}{-2}$ has direction ratios $\vec d_1=(2,1,-2)$.
2. The second line $\dfrac{x-1}{4}=\dfrac{y+1}{-4}=\dfrac{z-2}{2}$ has direction ratios $\vec d_2=(4,-4,2)$.
3. The angle $\theta$ between the lines satisfies $\cos\theta=\dfrac{\vec d_1\cdot\vec d_2}{|\vec d_1||\vec d_2|}$. …

Writing both planes with the same normal direction and applying the parallel-plane distance formula gives $\dfrac{\sqrt7}{2\sqrt2}$.

1. The two planes are $x+2y+3z+7=0$ and $2x+4y+6z+7=0$.
2. Divide the second equation by $2$ so both planes share the same coefficients for $x,y,z$: $2x+4y+6z+7=0 \;\Rightarrow\; x+2y+3z+\dfrac72=0$.
3. Now both planes have the form $x+2y+3z+d=0$, with $d_1=7$ for the first and $d_2=\dfrac72$ for the second — confirming the planes are parallel (same normal vector $(1,2,3)$).
4. The distance between two parallel planes $ax+by+cz+d_1=0$ and $ax+by+cz+d_2=0$ is $\dfrac{|d_1-d_2|}{\sqrt{a^2+b^2+c^2}}$.
5. Here $|d_1-d_2|=\left|7-\dfrac72\right|=\dfrac72$, and $\sqrt{a^2+b^2+c^2}=\sqrt{1^2+2^2+3^2}=\sqrt{14}$. …