Three Dimensional Geometry Test - 36

Direction cosine of $$a\hat{i}+b\hat{j}+c\hat{k}$$ can be given as $$\dfrac{a}{\sqrt{a^2+b^2+c^2}}, \dfrac{b}{\sqrt{a^2+b^2+c^2}}\dfrac{c}{\sqrt{a^2+b^2+c^2}}$$
Thus, here we have directional cosines as $$\dfrac{1}{\sqrt{1^2+1^2+(\sqrt 2)^2}}, \dfrac{1}{\sqrt{1^2+1^2+(\sqrt 2)^2}}, \dfrac{\sqrt 2}{\sqrt{1^2+1^2+(\sqrt 2)^2}}$$
=$$\displaystyle \frac {1}{2} , \frac {1} {2} , \frac {1} {\sqrt 2} $$

Let the coordinates of the foot of perpendicular $$P$$ from the origin to the plane be $$({x}_{1}, {y}_{1}, {z}_{1})$$
$$3y+4z-6=0$$
$$\Rightarrow$$ $$0x+3y+4z=6=6....(1)$$
The direction ratios of the normal are $$0,3$$ and $$4$$
$$\therefore$$ $$\sqrt { 0+{ \left( 3 \right)  }^{ 2 }+{ \left( 4 \right)  }^{ 2 } } =5$$
Dividing both sides of equation (1) by $$5$$, we obtain
$$0x+\cfrac{3}{5}y+\cfrac{4}{5}z=\cfrac{6}{5}$$
This equation is of the form $$lx+my+nz=d$$, where $$l,m,n$$ are the direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are
$$\left( 0,\cfrac { 3 }{ 5 } .\cfrac { 6 }{ 5 } ,\cfrac { 4 }{ 5 } .\cfrac { 6 }{ 5 }  \right) $$ i.e., $$\left( 0,\cfrac { 18 }{ 25 } ,\cfrac { 24 }{ 25 }  \right) $$

Let the coordinates of the foot of perpendicular $$O$$ from the origin to the plane be $$({x}_{1}, {y}_{1}, {z}_{1})$$
$$2x+3y+4z-12=0$$
$$\Rightarrow$$ $$2x+3y+4z=12......(1)$$
The direction ratios of normal are $$2,3$$ and $$4$$
$$\therefore$$ $$\sqrt { { \left( 2 \right)  }^{ 2 }+{ \left( 3 \right)  }^{ 2 }+{ \left( 4 \right)  }^{ 2 } } =\sqrt { 29 } $$
Dividing both sides of equation (1) by $$\sqrt {29}$$, we obtain
$$\cfrac { 2 }{ \sqrt { 29 }  } x+\cfrac { 3 }{ \sqrt { 29 }  } y+\cfrac { 4 }{ \sqrt { 29 }  } z=\cfrac { 12 }{ \sqrt { 29 }  } $$
This equation is of the form $$lx+my+nz=d$$, where $$l,m,n$$ are the direction cosines of normal to the plane and $$d$$ is the distance of normal from the origin.
The coordinates of the foot of the perpendicular are given by $$(ld,md,nd)$$
Therefore, the coordinates of the foot of the perpendicular are
$$\left( \cfrac { 2 }{ \sqrt { 29 }  } \cdot\cfrac { 12 }{ \sqrt { 29 }  } ,\cfrac { 3 }{ \sqrt { 29 }  } .\cfrac { 12 }{ \sqrt { 29 }  } ,\cfrac { 4 }{ \sqrt { 29 }  } \cdot\cfrac { 12 }{ \sqrt { 29 }  }  \right) $$ i.e., $$\left( \cfrac { 24 }{ 29 } ,\cfrac { 36 }{ 29 } ,\cfrac { 48 }{ 29 }  \right) $$

If the line $$\vec{OR}$$ makes angles $$\theta_1, \theta_2, \theta_3$$ with the planes $$XOY, YOZ, ZOX$$ respectively, then $$\cos^2\theta_1+\cos^2\theta_2+\cos^2\theta_3$$ $$=1$$

The direction ratios of the line segment joining $$A(-3, 1, 2)$$ and $$B(1, 4, -10)$$ are proportional to $$1-(-3), 4-1, -10-2,$$ i.e., $$4, 3, -12$$.
Now, its direction cosines are $$\displaystyle\frac{4}{\sqrt{4^2+3^2+(-12)^2}}, \displaystyle\frac{3}{\sqrt{4^2+3^2+(-12)^2}}, \displaystyle\frac{-12}{\sqrt{4^2, 3^2+(-12)^2}}$$
or $$\displaystyle\frac{4}{13}, \frac{3}{13}, -\frac{12}{13}$$

Given expression, $$\sin^2\alpha+\sin^2\beta+\sin^2\gamma$$
$$=(1-\cos^2\alpha)+(1-\cos^2\beta)+(1-\cos^2\gamma)$$
$$=3-\cos^2\alpha+\cos^2\beta+\cos^2\gamma=3-1=2$$
$$(\because \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1)$$

Since, the points $$A(1,6), B(3,-4)$$ and $$C(x,y)$$ are colinear
$$\therefore$$ $$\begin{vmatrix} 1 & 6 & 1 \\ 3 & -4 & 1 \\ x & y & 1 \end{vmatrix}=0$$
$$\Rightarrow $$ $$1(-4-y)-6(3-x)+1(3y+4x)=0$$
$$\Rightarrow $$ $$10x+2y-22=0$$
$$\Rightarrow $$ $$5x+y-11=0$$

$$\begin{vmatrix}k-1&k+2&1\\k&k+1&1\\k+1&k&1 \end{vmatrix}=0$$