How to solve the problem of distance between straight lines, distance from point to plane and distance between two parallel planes by vector calculation

How to solve the problem of distance between straight lines, distance from point to plane and distance between two parallel planes by vector calculation

It is suggested that the geometric method and the vector method be used to find the distance between the lines of different planes as follows:1(definition method) to find the length of the common vertical line segment of the lines of different planes;2(volume method) to find the height of the geometry;3(transformation method) to find the distance between parallel lines or parallel planes;4(maximum method) to construct the function of the distance between two points on the lines of different planes, but...

It is suggested that the geometric method and the vector method be used to find the distance between the lines of different planes as follows:1(the definition method) to find the length of the common vertical line segment of the lines of different planes;2(the volume method) to find the height of the geometry;3(the transformation method) to find the distance between parallel lines or parallel planes;4(the maximum method) to construct the function of the distance between two points on the lines of different planes, but...

How to find the distance between two straight lines with different planes by space vector in solid geometry How to find the distance of two straight lines with different planes by space vector in solid geometry

As shown in the figure, the and are known, any point on the straight line where the is located and
A vector formed by connecting lines at any point on a line, and
Normal vector of, d is the distance between two opposite lines of and , then
Because d=|| cosθ, cosθ=|•|/||||
So d =|•|/||(d is equal to
The quantity product is divided by the norm of the normal vector, and the formula is simplified as follows:" d equals, the absolute bridge product is divided by the norm ",
And the bridge vector is named for its nature, it's like a bridge,
Connect and together, so this name)
Last but not least, is how to find the normal vector
Because =(x1, y1, z1),=(x2, y2, z2)
So =(+(z2*y1-y2*z1),-(z2*x1-x2*z1),+(y2*x1-x2*y1))
The derivation process is omitted. In case of doubt, the can be calculated with and respectively.
Is it all 0, or can you deduce this conclusion by yourself
Note: As the vector expression is inconvenient, so replace with < >, please forgive!

Proving the formula of point-to-line distance Proving the formula of distance from point to straight line

The formula of distance from point to straight line is derived as follows: For point P (x0, y0), make PQ vertical straight line Ax+By+C=0, make PM parallel Y-axis at Q, make X-line at M; make PN parallel X-axis at N, make M (x1, y1) x1=x0, y1=(-Ax0+C)/B.PM=|y0-y1|=|y0+(Ax0+C)/B|=|(Ax0+By0+C)/B|, make N (x2, y2).y.

Let m and n be two unit vectors with an angle of 60°. Try to find the angle between a=2m+n and b=2n-3m. Such as title Although baidu has the answer, but I can not understand Are there any steps that I can understand...

Calculate the module |a| of a, the module |b| of b and the product a*b of a and b, and obtain the included angle θ from cosθ=(a*b)/(|a*|b|)
M*n=1/2, m*m=1, n*n=1
|^2=(2M+n)*(2m+n)=4(m*m)+n*n+4(m*n)=7, so | a|=√7
|B|^2=(2n-3m)*(2n-3m)=9(m*m)+4(n*n)-12(m*n)=7, so |b |=√7
A*b=(2m+n)*(2n-3m)=-6(m*m)+2(n*n)+m*n=-7/2
So cosθ=-1/2,θ=120°

Let m and n be two unit vectors with an included angle of 60. Try to find the included angle a=2m+n and b=2n-3m. Let m and n be two unit vectors with an angle of 60. Try to find the angle between vector a=2m+n and b=2n-3m. Let m and n be two unit vectors with an included angle of 60. Try to find the included angle of vector a=2m+n and b=2n-3m.

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4. If the included angle unit vector a and vector b is 60°, then |vector a-vector b|=

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