Consider two lines in space having equations r-=2i^+j^−k^+λ(i^−2j^+k^), r-=83i^−3j^+k^+μ(2i^−j^+k^) A line l1 from origin intersects the given lines. Find equation of projection line of l1 in plane x – y +2z = 0

x−21=y−1−2=z+11=λ, x−832=y+3−1=z−11=μ Equation required line x5=y−5=z2=5⇒x+y=0⇒2y+5z=0Plane containing linex+y+λ(2y+5z)=0⇒x+(2λ+1)y+5λz)→(1)x−y+23=0 is ⊥ to →(1)1−(2λ+1)+10λ=01−2λ−1+10λ=0⇒λ=0x+y=0x+y=0=x−y+23 D.R. of linei^j^k^1101−12=i^(2)−j^(2)+k^(−1−1)⇒2i^−2j^−2k^ ⇒2(i^−j^−k^) Equation of line is x1=y−1=z−1

Consider two lines in space having equations r-=2i^+j^−k^+λ(i^−2j^+k^), r-=83i^−3j^+k^+μ(2i^−j^+k^) A line l1 from origin intersects the given lines. Find equation of projection line of l1 in plane x – y +2z = 0

x−21=y−1−2=z+11=λ, x−832=y+3−1=z−11=μ Equation required line x5=y−5=z2=5⇒x+y=0⇒2y+5z=0Plane containing linex+y+λ(2y+5z)=0⇒x+(2λ+1)y+5λz)→(1)x−y+23=0 is ⊥ to →(1)1−(2λ+1)+10λ=01−2λ−1+10λ=0⇒λ=0x+y=0x+y=0=x−y+23 D.R. of linei^j^k^1101−12=i^(2)−j^(2)+k^(−1−1)⇒2i^−2j^−2k^ ⇒2(i^−j^−k^) Equation of line is x1=y−1=z−1

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Consider two lines in space having equations $\bar{r} = 2 \hat{i} + \hat{j} - \hat{k} + λ (\hat{i} - 2 \hat{j} + \hat{k}), \bar{r} = \frac{8}{3} \hat{i} - 3 \hat{j} + \hat{k} + μ (2 \hat{i} - \hat{j} + \hat{k})$ A line $l_{1}$ from origin intersects the given lines. Find equation of projection line of $l_{1}$ in plane x – y +2z = 0

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x = − y = 2 z

x = y = z

x = − y = z

− x = y = z

answer is B.

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Detailed Solution

$\frac{x - 2}{1} = \frac{y - 1}{- 2} = \frac{z + 1}{1} = λ, \frac{x - \frac{8}{3}}{2} = \frac{y + 3}{- 1} = \frac{z - 1}{1} = μ$
Equation required line
$\frac{x}{5} = \frac{y}{- 5} = \frac{z}{2} = 5 \Rightarrow x + y = 0 \Rightarrow 2 y + 5 z = 0$
Plane containing line
$\begin{array}{l} x + y + λ (2 y + 5 z) = 0 \Rightarrow x + (2 λ + 1) y + 5 λ z) \to (1) \\ x - y + 23 = 0 is ⊥ to \to (1) 1 - (2 λ + 1) + 10 λ = 0 \\ 1 - 2 λ - 1 + 10 λ = 0 \Rightarrow λ = 0 \\ x + y = 0 \\ x + y = 0 = x - y + 23 \end{array}$

D.R. of line $|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & - 1 & 2 \end{array}| = \hat{i} (2) - \hat{j} (2) + \hat{k} (- 1 - 1) \Rightarrow 2 \hat{i} - 2 \hat{j} - 2 \hat{k}$
$\Rightarrow 2 (\hat{i} - \hat{j} - \hat{k}) Equation of line is \frac{x}{1} = \frac{y}{- 1} = \frac{z}{- 1}$