Find the shortest distance between the line r = 2î - 2ĵ + 3k̂ + λ(î - ĵ + 4k̂) and the plane r · (î + 5ĵ + k̂) = 5.

Question

Accepted Answer

(10)/(3√(3)). The shortest distance from a line to a plane is found by projecting a vector from a point on the line to the plane along the direction normal to the plane. In this case, the normal vector of the plane is ( extbf{n} = extbf{i} + 5 extbf{j} + extbf{k}), and the distance can be calculated using the formula that incorporates the line's direction and the plane's equation, resulting in the correct answer of ( rac{10}{3 ext{√}3}).

Find the shortest distance between the line r = 2î - 2ĵ + 3k̂ + λ(î - ĵ + 4k̂) and the plane r · (î + 5ĵ + k̂) = 5.

Solution

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