Question #188589

Let L1 be the line in R3 with equation (x,y,z)=(1,0,2)+t(−1,3,4); t∈R

and let L2 be the line that is parallel to L1 and contains the point (1, −1, 3). Let V be the plane that contains both the lines L1 and L2.

(a) Find two vectors that are both parallel to the plane V but are not parallel to one another.

(b) Find a vector that is perpendicular to the plane V .

(c) Find an equation for the plane V .

(d) Find an equation for the line L3 that is perpendicular to the plane V and contains the point (1, −1, 4) .

Hint: Find a parametric equation for L3. Don’t try to find a Cartesian equation for L3.


Expert's answer

L1:(x,y,z)=(1,0,2)+t(−1,3,4),t∈RL1∣∣L2L_1 : (x,y,z)=(1,0,2)+t(-1,3,4), t\in R\\ L_1 ||L_2 \\


L1:(x,y,z)=(1,−1,3)+t(−1,3,4),t∈RV:L1∈V,L2∈VL_1 : (x,y,z)=(1,-1,3)+t(-1,3,4), t\in R\\ V: L_1\in V, L_2\in V


a)

a⃗=(−1,3,4)∣∣VA(1,0,2)∈L1B(1,−1,3)∈L2\vec{a}=(-1,3,4)||V\\ A(1,0,2)\in L_1\\ B(1,-1,3)\in L_2\\


AB⃗=(1−1,−1−0,3−2)=(0,−1,1)∣∣Va⃗∤∣AB⃗−10≠3−1≠41\vec{AB}=(1-1,-1-0,3-2)=(0,-1,1)|| V\\ \vec{a}\not|| \vec{AB}\\ \frac{-1}{0}\not=\frac{3}{-1}\not=\frac{4}{1}


b)

n⃗=[a⃗,AB⃗]=∣i⃗j⃗k⃗−1340−11∣==i⃗(3+4)−j⃗(−1−0)+k⃗(1−0)=(7,1,1)\vec{n}=[\vec{a},\vec{AB}]=\begin{vmatrix} \vec{i} & \vec{j}&\vec{k} \\ -1&3&4\\ 0&-1&1 \end{vmatrix}=\\ =\vec{i}(3+4)-\vec{j}(-1-0)+\vec{k}(1-0)=(7,1,1)

n⃗\vec{n} is perpendicular to the plane VV



c)

n⃗=(7,1,1)\vec{n}=(7,1,1) is perpendicular to the plane, A(1,0,2)∈VA(1,0,2)\in V

7(x−1)+1(y−0)+1(z−2)=07x+y+z−9=07(x-1)+1(y-0)+1(z-2)=0\\ 7x+y+z-9=0



d)

L3L_3 is perpendicular to the plane VV

n⃗=(7,1,1)∣∣L3C(1,−1,4)∈L3(x,y,z)=(1,−1,4)+t(7,1,1),t∈R\vec{n}=(7,1,1)||L_3\\ C(1,-1,4)\in L_3\\ (x,y,z)=(1,-1,4)+t(7,1,1), t\in R



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