Three Dimensional Geometry

The direction vector for the line joining points $(1,2,3)$ and $(-1,4,2)$

$ A=(-1-1,4-2,2-3)=(-2,2,-1) $

The direction vector for the line joining points $(-2,3,3)$ and $(0,6,-3)$.

$ \mathbf{B}=(0-(-2), 6-3,-3-3)=(2,3,-6) $

Now, $\mathbf{A} \cdot \mathbf{B}=(-2)(2)+(2)(3)+(-1)(-6)$

$ =-4+6+6=8 $

And, $|\mathbf{B}|=\sqrt{(2)^2+(3)^2+(-6)^2}$

$ =\sqrt{4+9+36}=\sqrt{49}=7 $

The magnitude of projection of vector $\mathbf{A}$ onto $\mathbf{B}$ is given by

$ \left|\frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{B}|}\right| $

$ \text { Thus, the magnitude of projection }=\left|\frac{8}{7}\right|=\frac{8}{7} $

$\text { Any point on the line } \frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1} \text {, is of the form } (-3 \lambda-1,2 \lambda+3,-\lambda+2).$

Let the point $A \equiv(-3 \lambda-1,2 \lambda+3,-\lambda+2)$ and $B \equiv(-4,3,1)$

Then, $\mathbf{A B}=(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}}$

Now, as the line is parallel to the given plane, therefore $\mathrm{AB}$ will be parallel to the given plane and so $\mathrm{AB}$ will be perpendicular to the normal of plane.

$\Rightarrow \mathrm{AB} \cdot \mathrm{n}=0$, where $\mathrm{n}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathbf{k}}$ is normal to the plane.

$\begin{aligned} & \Rightarrow(3 \lambda-3) \hat{\mathbf{i}}-2 \lambda \hat{\mathbf{j}}+(\lambda-1) \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})=0 \\ & \Rightarrow 3 \lambda-3-4 \lambda-\lambda+1=0 \\ & \Rightarrow-2 \lambda=2 \Rightarrow \lambda=-1 \end{aligned}$

Now, the required equation is the equation of line joining $A(2,1,3)$ and $B(-4,3,1)$, which is

$\begin{array}{ll} & \frac{x-(-4)}{2-(-4)}=\frac{y-3}{1-3}=\frac{z-1}{3-1} \\ \Rightarrow & \frac{x+4}{6}=\frac{y-3}{-2}=\frac{z-1}{2} \\ \text { or } & \frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1} \end{array}$

Given plane $3x + y + 2z + 6 = 0$

and line ${{x - {1 \over 3}} \over {{{2b} \over 3}}} = {{y - 3} \over { - 1}} = {{z - 1} \over a}$

Since, plane is parallel to line, then

$3\left( {{{2b} \over 3}} \right) + (1)( - 1) + 2(a) = 0$

$2b - 1 + 2a = 0$

$ \Rightarrow a + b = {1 \over 2}$

Now, $3a + 3b = {3 \over 2}$

Let each edge of cube be a, then coordinates of the vertices of cube are

O(0, 0, 0), A(a, 0, 0), B(0, a, 0), C(0, 0, a), N(a, a, 0), P(a, a, a), L(0, a, a), M(a, 0, a)

Direction ratios of the diagonals OP, AL, BM and CN are (a, a, a), ($-$a, a, a), (a, $-$a, a) and (a, a, $-$a).

Let $\theta$ be the acute angle between diagonals OP and AL.

$\therefore$ $\cos \theta = {{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}$

$ = {{a \times ( - a) + a \times a + a \times a} \over {\sqrt {{a^2} + {a^2} + {a^2}} \sqrt {{{( - a)}^2} + {a^2} + {a^2}} }}$

$ = {{ - {a^2} + {a^2} + {a^2}} \over {\sqrt {3{a^2}} \sqrt {3{a^2}} }} = {{{a^2}} \over {a\sqrt 3 \times a\sqrt 3 }}$

$ \Rightarrow \cos \theta = {1 \over 3} \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over 3}} \right)$

Given lines are

${L_1}:{{x + 1} \over 3} = {{y + 2} \over 1} = {{z + 1} \over 2}$ and ${L_2}:{{x - 2} \over 1} = {{y + 2} \over 2} = {{z - 3} \over 3}$

Now, convert into vector form

${L_1}:( - \widehat i - 2\widehat j - \widehat k) + \lambda (3\widehat i + \widehat j + 2\widehat k)$

${L_2}:(2\widehat i + 2\widehat j + 3\widehat k) + \mu (\widehat i + 2\widehat j + 3\widehat k)$

Line L₁ comparing with ${a_1} + \lambda {b_1}$

and L₂ comparing with ${a_2} + \mu {b_2}$, then we have ${b_1} = 3\widehat i + \widehat j + 2\widehat k$ and ${b_2} = \widehat i + 2\widehat j + 3\widehat k$

Perpendicular to both b₁ and b₂ = ${b_1} \times {b_2}$

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 1 & 2 \cr 1 & 2 & 3 \cr } } \right|$

$ = \widehat i(3 - 4) - \widehat j(9 - 2) + \widehat k(6 - 1) = - \widehat i - 7\widehat j + 5\widehat k$

$\therefore$ Required unit vector $ = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {{{( - 1)}^2} + {{( - 7)}^2} + {{(5)}^2}} }}$

$ = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {\sqrt {75} }} = {{ - \widehat i - 7\widehat j + 5\widehat k} \over {5\sqrt 3 }}$

The given line is r = a + tb, where $a = 2\widehat i - 2\widehat j + 3\widehat k$, $b = \widehat i - \widehat j + 4\widehat k$ and given plane is r . n = d, where $n = \widehat i + 5\widehat j + \widehat k$, d = 5.

Since, $b\,.\,n = 1 - 5 + 4 = 0$

$\therefore$ Given line is parallel to given plane.

$\therefore$ The distance between the line and the plane is equal to length of the perpendicular from the point a = ${2\widehat i - 2\widehat j + 3\widehat k}$ on the line to the given plane.

$\therefore$ Required distance $ = \left| {{{(2\widehat i - 2\widehat j + 3\widehat k)\,.\,(\widehat i + 5\widehat j + \widehat k) - 5} \over {\sqrt {1 + 25 + 1} }}} \right|$

$ = \left| {{{2 - 10 + 3 - 5} \over {\sqrt {27} }}} \right| = {{10} \over {3\sqrt 3 }}$

We have P(3, 7, 1) and R(2, 5, 7)

DR's of PR = ($-$1, $-$2, 6)

Angle between line PQ and plane is

$\sin \theta = {{( - 1)(3) + ( - 2) \times 2 + 6 \times 11} \over {\sqrt {1 + 4 + 36} \sqrt {9 + 4 + 121} }}$

$\sin \theta = {{59} \over {\sqrt {41} \sqrt {134} }}$

$PM = \left| {{{3(3) + 2(7) + 11( \times 1) - 9} \over {\sqrt {9 + 4 + 121} }}} \right|$

$ = \left| {{{9 + 14 + 11 - 9} \over {\sqrt {134} }}} \right| = {{25} \over {\sqrt {134} }}$

$\sin \theta = {{PM} \over {PQ}}$

$ \Rightarrow PQ = {{PM} \over {\sin \theta }} = {{25} \over {\sqrt {134} }} \times {{\sqrt {41} \times \sqrt {134} } \over {59}} = {{25\sqrt {41} } \over {59}}$