Relation between planes a₁x + b₁y + c₁z + d₁ = 0 and a₂x + b₂y + c2₁z + d₂ = 0, if ir normal are parallel to each or is a₁ : b₁ : c₁ = a₂ : b₂ : c₂ ⇒ \(\frac {a1}{a2} = \frac{b1}{b2} = \frac{c1}{c2}\).

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

Given that equations of lines are
\(\vec{r}=\vec{a_1}+?\vec{b_1} \,and \,\vec{r}=\vec{a_2}+?\vec{b_2}\)
? angle between two lines is given by
cos??=\(\left |\frac{\vec{b_1}.\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right |\).

If three points are collinear, n y will be in a straight line and hence determinant of three points will be zero.
i.e.\(\begin{vmatrix}x_1&y_1&z_1\\x_2&y_2&z_2\\x_3&y_3&z_3\end{vmatrix}\)=0

Given that, normal to planes are \(\vec{n_1}=2\hat{i}-\hat{j}+\hat{k}\) and \(\vec{n_2}=3\hat{i}+2\hat{j}-3\hat{k}\)
cos??=\(\left |\frac{\vec{n_1}.\vec{n_2}}{|\vec{n_1}||\vec{n_2}|}\right |\)
\(|\vec{n_1}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{6}\)
\(|\vec{n_2}|=\sqrt{3^2+2^2+(-3)^2}=\sqrt{22}\)
\(\vec{n_1}.\vec{n_2}\)=2(3)-1(2)+1(-3)=6-2-3=1
cos??=\(\frac{1}{\sqrt{22}.\sqrt{6}}=\frac{1}{\sqrt{132}}\)
??=\(cos^{-1}?\frac{1}{\sqrt{132}}\).

&ta; = 90 degrees
The relation between plane ax + by + cz + d = 0 and a₁, b₁, c₁ direction ratios of a line, if plane and line are perpendicular to each or is \(\frac {a}{a1} = \frac{b}{b1} = \frac{c}{c1}\).

The relation between plane ax + by +cz + d = 0 and a₁, b₁, c₁ direction ratios of a line, if plane and line are parallel to each or is a₁a + b₁b + c₁c = 0.

We know that, shortest distance between two skew lines is given by
d=\(\left |\frac{(\vec{b_1}�\vec{b_2}).(a_2-a_1)}{|\vec{b_1}�\vec{b_2}|}\right |\)
The vector equations of two lines is
\(\vec{r}=2\hat{i}+2\hat{j}-2\hat{k}+?(3\hat{i}+2\hat{j}+5\hat{k})\)
\(\vec{r}=4 \hat{i}-\hat{j}+5\hat{k}+?(3\hat{i}-2\hat{j}+4\hat{k})\)
?d=\(\left|\frac{((3\hat{i}+2\hat{j}+5\hat{k})�(3\hat{i}-2\hat{j}+4\hat{k}).(4\hat{i}-\hat{j}+5\hat{k})-(2\hat{i}+2\hat{j}-2\hat{k}))}{|(3\hat{i}+2\hat{j}+5\hat{k})�(3\hat{i}-2\hat{j}+4\hat{k})|}\right |\)
\((3\hat{i}+2\hat{j}+5\hat{k})�(3\hat{i}-2\hat{j}+4\hat{k})=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&5\\3&-2&4\end{vmatrix}\)
=\(\hat{i}(8+10)-\hat{j}(12-15)+\hat{k}(-6-6)\)
=\(18\hat{i}+3\hat{j}-12\hat{k}\)
d=\(\left|\frac{(18\hat{i}+3\hat{j}-12\hat{k}).(2\hat{i}-3\hat{j}+7\hat{k})}{\sqrt{18^2+3^2+(-12)^2}}\right |\)
d=\(\left|\frac{36-9-84}{\sqrt{477}}\right |\)=\(\frac{57}{\sqrt{477}}\).

Given that, a, b, c are direction ratios of line and l, m, n are direction cosines of line,
\(\frac{l}{a}=\frac{m}{b}=\frac{n}{c}=k\) and l²+m²+n²=1
?l=ak, m=bk, n=ck
(ak)²+(bk)²+(ck)²=1
k² (a²+b²+c²)=1
k²=\(\frac{1}{a^2+b^2+c^2}\)
?k=±\(\frac{1}{\sqrt{(a^2+b^2+c^2)}}\)
Hence, l²-m²=n²-1 is incorrect.

The angle between two lines is given by equation
\(cos??=\left |\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}}\right |\)
cos?90�=\(\left |\frac{3(1)+p(1)+1(-2)}{\sqrt{3^2+p^2+1^2}.\sqrt{1^2+1^2+(-2)^2}}\right |\)
0=\(|\frac{p+1}{\sqrt{10+p^2}.?6}|\)
0=p+1
p=-1