Tent. sudut yang lebih kecil dari perpotongan bidang 5x-14y+2z-8=0 dan 10x-11y+2z+5=0 (tent sudut antara normal-normalnya)
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"Find smallest angle of intersection of the plane of 5x-14y+2z-8=0 and 10x-11y+2z+5=0 (Find angle between the normals)".
Please check accuracy of translation.
The normal vectors are
(5,-14,2) and (10,-11,2).
We find the unit normal vectors:
(5,-14,2)/√(5²+(-14)²+2²)
=(1/15)(5,-14,2)
Similarly, the unit normal vector for (10,-11,2) is (1/15)(10,-11,2)
The inner product of the two unit vectors give the cosine of the angle between the normals:
(1/15)(1/15)[(5,-14,2)dot(10,-11,2)]
=(1/225)[50+154+4]
=208/225
cos-1(208/225)=22.4°.
Please check accuracy of translation.
The normal vectors are
(5,-14,2) and (10,-11,2).
We find the unit normal vectors:
(5,-14,2)/√(5²+(-14)²+2²)
=(1/15)(5,-14,2)
Similarly, the unit normal vector for (10,-11,2) is (1/15)(10,-11,2)
The inner product of the two unit vectors give the cosine of the angle between the normals:
(1/15)(1/15)[(5,-14,2)dot(10,-11,2)]
=(1/225)[50+154+4]
=208/225
cos-1(208/225)=22.4°.
1 answer