Tag: vectors: lines in two and three dimensions

$\begin{array}{l} let\, the\, point\, (5,4,-1)\, \, be\, \, P\, and\, the\, point\, through\, which\, the\, \, line\, passes\, \, be\, \, Q\, (1,0,0).\, \, \, \,  \ the\, line\, is\, parallel\, to\, the\, vector:\, \, \overrightarrow { r } =\left( { 2\hat { i } +9\hat { i } +5\hat { k }  } \right)  \ Now, \ \overrightarrow { PQ } =-4\hat { i } -4\widehat { j } +\hat { k }  \ \therefore \, \, \, \overrightarrow { r\,  } \, \times \overrightarrow { PQ } =\left| \begin{array}{l} \, \, \hat { i } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \widehat { j } \, \, \, \, \, \, \, \, \, \, \, \widehat { k }  \ \, \, 2\, \, \, \, \, \, \, \, \, \, \, \, \, \, 9\, \, \, \, \, \, \, \, \, \, \, 5\, \,  \ \, -4\, \, \, \, \, \, \, -4\, \, \, \, \, \, \, \, \, \, 1 \end{array} \right| \, \, \, \, \, \, \, \,  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\, 29\, \hat { i } \, -\, \, 22\, \widehat { j } \, +28\, \widehat { k }  \ \Rightarrow \left| { \, \overrightarrow { r\,  } \, \times \overrightarrow { PQ }  } \right| =\sqrt { \, { { (29) }^{ 2 } }\, +{ { (-\, \, 22) }^{ 2 } }\, +{ { (28) }^{ 2 } } }  \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =\sqrt { 841+484+784 } =\sqrt { 2109 }  \ \left| { \overrightarrow { r\,  } \,  } \right| =\sqrt { { 2^{ 2 } }+{ 9^{ 2 } }+{ 5^{ 2 } } }  \ \, \, \, \, \, \, \, =\sqrt { 4+81+25 } =\sqrt { 110 }  \ d=\frac { { \, \left| { \overrightarrow { r\,  } \, \times \overrightarrow { PQ }  } \right|  } }{ { \left| { \overrightarrow { r\,  } \,  } \right|  } } =\frac { { \sqrt { 2109 }  } }{ { \sqrt { 110 }  } } =\sqrt { \frac { { 2109 } }{ { 110 } }  }  \ so\, that\, the\, correct\, option\, is\, C. \end{array}$

$\dfrac{|d|}{\sqrt {a^2+b^2+c^2}}\\dfrac{1}{\sqrt {2^2+1^2+2^2}}\\dfrac 1{\sqrt 9}=\dfrac 13$

$P _1$ is perpendicular distance of A from XY-plane 

So, $x _1=0,y _1=0,z _=-3$

$\therefore p _1=\sqrt{(-3)^2}=3$

$p _2$ is perpendicular distance from Y-axis

So, $x _1=4,y _1=0,z _1=-3$

$p _{2} = \sqrt {x _{1}^{2} + y _{1}^{2}}$

$\therefore p _2=\sqrt{4^2+(-3^2)}=5$

$\vec r = \left( {3\hat i + 7\hat j + \hat k} \right) + t\left( {\hat j + \hat k} \right)$

 let the foot of the perpendicular from point +p to the given line be $\theta \left( {x{,^1}y{,^1},{z^1}} \right)$ then direction ratios of $pq$ are

$\left( {{x^1} - 5,{y^1} - 1,{z^1} - 3} \right)$

since $pq$is $ \bot $ to the given line so 

$b \times \left( {{x^1} - 5} \right) + 1 \times \left( {{y^1} - 11 + 1 \times \left( {{z^1} - 31} \right)} \right) = 0$

${y^1} + 1 + {z^1} - 3 = 0$

${y^1} + {z^1} = 4 = 0$ 

artesian equation of the given line is 

$\frac{{x - 3}}{0} = \frac{{y - 7}}{1} = \frac{{z - 1}}{1}$ 

let $\frac{{x - 3}}{0} = \frac{{y - 7}}{1} = \frac{{z - 1}}{1} = r$

so general point on this line is (3,r+7,r+1)

since $\left( {{x^1},{y^1},{z^1}} \right)$

lies on this line so ${x^1} = 3\,\,{y^1} = r + 7\,\,{z^1} = r + 1$

putting values of ${y^1}& $ in

${y^1} + {z^1} = 4$

$r + 7 + r + 1 = 4$

$2r=-4$

$=r=-2$

thus ${x^1} = 3,{y^1} =  - 2 + 7 = 5$

${z^1} =  - 2 + 1 =  - 1$

so point Q is $(3,5,-1)$

therefore perpendicular distance $\sqrt {{{\left( {5 - 3} \right)}^2} + {{\left( {1 - 5} \right)}^2} + {{\left( {3 + 1} \right)}^2}} $

$\sqrt {4 + 16 + 16} $

$\sqrt {36} $

$=6units$

Equation of the line joining $A$ and $B$ is 

$\dfrac{x-2}{3-2}=\dfrac{y-1}{-1-1}=\dfrac{z-2}{4-2}$

$\dfrac{{x - 2}}{1} = \dfrac{{y - 1}}{{ - 2}} = \dfrac{{z - 2}}{2}$

Let

$\dfrac{{x - 2}}{1} = \dfrac{{y - 1}}{{ - 2}} = \dfrac{{z - 2}}{2}=r$

$x=r+2,\,y=-2r+1,\,z=2r+2,\,z=2r+2$

General point on the given line is 

$(r+2,-2r+1,2r+2)$

Let the foot of perpendicular from $P$ to the given line be $Q(r _1+2,-2r _1+1,2r _1+2)$

Direction ratios the line $PQ$ are 

$(r _1+2-6,\,-2r _1+1+4,\,2r _1+2-4)$ which is 

$r _1-4,\,\,-2r _1+5,\,\,2r _1-2$

Since the line $PQ$ is perpendicular to the given line 

$1(r _1-2)-2(-2r _1+5)+(2r _1-2)=0$

$r _1=2$

$\begin{array}{l} { r _{ 1 } }+2=4 \  \ -2{ r _{ 1 } }+1=-3 \  \ 2{ r _{ 1 } }+2=6 \end{array}$

Therefore, 

Foot of perpendicular is $Q(4,-3,6)$

Perpendicular distance $=PQ$

$=\sqrt {(4-6)^2+(-3+4)^2+(6-4)^2}=3$

Points are $P(2,4,3), A(1,2,4), B(3,4,5).$ 

To find the foot of the perpendicular from $P$ onto $AB$.

Equation of $AB$

$x = 1+(3-1)t = 1+2t$

$y = 2+(4-2)t = 2+2t$

$z = 4+(5-4)t = 4+t$

Direction Coefficients

$ 2,  2,  1$ $Q$ lies on the $AB.$

Equation of a $3-d$ plane perpendicular to $AB$ 

$2x + 2y + z = C $

This plane passes through $P$

So,

 $2\times 2+2\times 4+3= C$

$Q$ lies on the line and the plane. 

So,

$4t+2 + 4t+4 + t+4 = 15$

So for $Q$ $t = 5/9$

Hence, $Q=(19/9, 28/9, 41/9)$

Hence, this is the answer.