3 years ago

Find the distance from the point to the given plane. (1, −7, 8), 3x + 2y + 6z = 5

1 answer:

5 0

3x + 2y + 6z = 5??
then
distance = |3(1) + 2(-2) + 6(4) - 5| / √(3^2 + 2^2 + 6^2)
the plane is ax + by + cz - d = 0 while the point is (x1, y1, z1)

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To show: the set $\{T(v_1), T(v_2), T(v_3)\}$ is linearly dependent.

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If $\{v_1,v_2,v_3,...,v_n\}$ is a set of linearly dependent vectors then there exists atleast one $k_i:i=1,2,3,...,n$ such that $k_1v_1+k_2v_2+k_3v_3+...+k_nv_n=0$

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$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0\\T(k_1v_1)+T(k_2v_2)+T(k_3v_3)=0\\T(k_1v_1+k_2v_2+k_3v_3)=0\\$

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$k_1T(v_1)+k_2T(v_2)+k_3T(v_3)=0$

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