Let l, m, n be direction cosines of line.
We know that, if ?, ?, ? are angles that line makes with x, y, z-axis respectively, n
l=cos??=cos?60°=\(\frac{1}{2}\)
m=cos??=cos?150°=-\(\frac{\sqrt{3}}{2}\)
n=cos??=cos?45°=\(\frac{1}{\sqrt{2}}\).

If l₁ and l₂ are two parallel lines, n y are coplanar and hence can be represented by following equations
\(\vec{r}=\vec{a_1}+?\vec{b}\)
\(\vec{r}=\vec{a_2}+?\vec{b}\)
Then distance between lines is given by
d=\(\left|\frac{\vec{b}×(\vec{a_2}-\vec{a_1})}{|\vec{b}|}\right |\)

The given statement is true.
We know that angle between two lines is given by formula
cos??=\(\left |\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |\)
So, if lines L₁ and L₂ are perpendicular to each to each or n,
?=90°
?\(a_1 \,a_2+b_1 \,b_2+c_1 \,c_2\)=0

Relation between planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c2₁z + d₂ = 0, if ir normal are perpendicular to each or is a₁a₂ + b₁b₂ + c₁c₂ = 0.
1(3) + 2(4) + k(-1) = 0
k(-1) = -11
k = 11

Let \(\vec{a}=\hat{i}+2\hat{j}-\hat{k}, \,\vec{b}=-\hat{j}+2\hat{k}, \,\vec{c}=3\hat{i}+\hat{j}+\hat{k}\)
The vector equation of plane passing through three points is given by
\((\vec{r}-\vec{a}).[(\vec{b}-\vec{a})×(\vec{c}-\vec{a})]\)=0
\((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[((-\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-\hat{k}))×((3\hat{i}+\hat{j}+\hat{k})?(\hat{i}+2\hat{j}-\hat{k}))]\)=0
\((\vec{r}-(\hat{i}+2\hat{j}-\hat{k})).[(\hat{i}-3\hat{j}+3\hat{k})×(2\hat{i}-\hat{j}+2\hat{k})]\)=0

Angle between a plane and a line sin &ta;=\(\frac {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\)
sin&ta; = 0.49
&ta; = sin^-1(0.49)
&ta; = 29.34

The above equations can also be expressed as
\(\vec{r}=(\hat{i}-3\hat{j}+\hat{k})+p(-\hat{i}+\hat{j}+\hat{k})\)
\(\vec{r}=(-\hat{i}-3\hat{j}+\hat{k})+q(\hat{i}-2\hat{j}+\hat{k})\)
The distance between two lines is given by
d=\(\left|\frac{(\vec{b_1}×\vec{b_2}).(a_2-a_1)}{|\vec{b_1}×\vec{b_2}|}\right |\)
=\(\left|\frac{((-\hat{i}+\hat{j}+\hat{k})×(\hat{i}-2\hat{j}+\hat{k})).(-\hat{i}-3\hat{j}+\hat{k})-(\hat{i}-3\hat{j}+\hat{k})}{|-\hat{i}+\hat{j}+\hat{k})×(\hat{i}-2\hat{j}+\hat{k}|}\right |\)
\((-\hat{i}+\hat{j}+\hat{k})×(\hat{i}-2\hat{j}+\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&1\\1&-2&1\end{vmatrix}\)
=\(\hat{i}(1+2)-\hat{j}(-1-1)+\hat{k}(2-1)\)
=\(3\hat{i}+2\hat{j}+\hat{k}\)
d=\(\left|\frac{(3\hat{i}+2\hat{j}+\hat{k}).(-2\hat{i})}{\sqrt{3^2+2^2+1^2}}\right |=\frac{6}{\sqrt{14}}\)

We know that if scalar or dot product of vectors is 0, n y are at right angles to each or. Here dot product of normal vectors of plane is 0, i.e. \(\vec{n_1}.\vec{n_2}\)=0. Hence, planes will be perpendicular to each or.

The planes which are perpendicular to each or i.e.; having an angle 90 degrees between m are called orthogonal planes. Hence, Orthogonal planes have an angle 90 degrees between m.

The angle between a line and a plane is complement of angle between line L and a normal line to plane π. If &ta; is angle between line whose ratios are a₁, b₁, c₁ and plane ax + by + cz + d = 0 n sin &ta;=\(\frac {a1a+b1b+c1c}{\sqrt {a^2+b^2+c^2} \sqrt{a1^2+b1^2+c1^2 }}\).