How do I find the angle between the planes 5(x+1) + 3(y+2) + 2z = 0 and x + 3(y-1) + 2(z+4) = 0?

Question

Chloe Alexander · Accepted Answer

#~~38.69^@#

Let the given planes be #pi_1 and pi_2#. Let #vecn_1 and vec n_2#

be their normals.

Recall that for a plane #: ax+by+cz+d=0#, its normal #vecn=(a,b,c)#.

Clearly, #vec n_1=(5,3,2), and, n_2=(1,3,2)#

the #/_alpha# btwn. #pi_1 and pi_2# is, by defn.,

# alpha=arc cos{|vecn_1*vecn_2|/(||vecn_1||||vecn_2||)}#.

#=arc cos{|5*1+3*3+2*2|/((sqrt(25+9+4)sqrt(1+9+4))}}#

#=arc cos(18/(sqrt(2*19*2*7)))#

#=arc cos(9/sqrt(133))#

#=arc cos(9/11.53)#

#~~arc cos (0.7806)#

#~~38.69^@#

Serenity Johnson · Answer

undefined To find the angle between two planes, we can use the formula:

cos(theta) = |a₁ * a₂ + b₁ * b₂ + c₁ * c₂| / sqrt(a₁² + b₁² + c₁²) * sqrt(a₂² + b₂² + c₂²)

Where (a₁, b₁, c₁) and (a₂, b₂, c₂) are the normal vectors of the planes.

For the first plane, the coefficients of x, y, and z are (5, 3, 2), and for the second plane, the coefficients are (1, 3, 2).

Using these coefficients, we can calculate the angle between the planes by substituting them into the formula:

cos(theta) = |5 * 1 + 3 * 3 + 2 * 2| / sqrt(5² + 3² + 2²) * sqrt(1² + 3² + 2²)

cos(theta) = |5 + 9 + 4| / sqrt(38) * sqrt(14)

cos(theta) = |18| / sqrt(38 * 14)

cos(theta) = 18 / sqrt(532)

cos(theta) ≈ 0.983

Now, to find the angle, we take the inverse cosine of 0.983:

theta ≈ arccos(0.983)

theta ≈ 10.08 degrees

So, the angle between the planes is approximately 10.08 degrees.