Let \(\vec { \mu } \)  the position vector of an arbitrary point on the plane \(\vec { r } .(2\hat { i } -\hat { j } -2\hat { k } )\) = 6. Then, we have
\(\vec { \mu } .(2\hat { i } -\hat { j } -2\hat { k } )=6\) .........(1)
If δ is the distance between the given planes, then δ is the perpendicular distance from  \(\vec { \mu } \) to the plane
\(\vec { r } .(6\hat { i } -3\hat { j } -6\hat { k} )\) = 27
Therefore, δ = \(\frac { \left| \vec { u } .\vec { n } -p \right| }{ \left| \vec { n } \right| } =\left| \frac { \vec { u } .(6\hat { i } -3\hat { j } -6\hat { k } )-27 }{ \sqrt { 6^{ 2 }+(-{ 3) }^{ 2 }+(-{ 6 })^{ 2 } } } \right| =\left| \frac { 3(\vec { u } .(2\hat { i } -\hat { j } -2\hat { k } ))-27 }{ 9 } \right| =\left| \frac { (3(6)-27 }{ 9 } \right| =1\) unit

If the position vector of the given point is \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \), then the equation of the plane passing through a point and normal to a vector is given by \((\vec { r } -\vec { a } ).\vec { n } =0\) or \(\vec { r } .\vec { n } =\vec { a } .\vec { n } \)
Substituting \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \) in the above equation, we get
\(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )=(4\hat { i } +2\hat { j } -3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
Thus, the required vector equation of the plane is \(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )\)= 3. If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) then
we get the Cartesian equation of the plane 2x − y + z = 3.

Let \(\vec { a } =4\hat { i } +3\hat { j } -7\hat { k } \) and \(\vec { b } =2\hat { i } -6\hat { j } +7\hat { k } \)
Non-parametric form of vector equation of a straight line passing through a point (\(\vec {a} \)) and parallel to a vector (\(\vec { b } \)) is \((\vec { r } -\vec { a } )\times \vec { b } \)
⇒ \([\vec { r } -(4\hat { i } +3\hat { j } -7\hat { k } )]\times (2\hat { i } -6\hat { j } +7\hat { k } )=\vec { 0 } \)
Its cartesian equation is
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
⇒ \(\frac { x-4 }{ 2 } =\frac { y-3 }{ -6 } =\frac { z+7 }{ 7 } \)
[∵ (x₁, y₁, z₁) is (4, 3, -7) & (b₁, b₂, b₃) is (2, -6, 7)]

We get \([\vec { a } +\vec { c } ,\vec { a } +\vec { b } ,\vec { a } +\vec { b } +\vec { c } ]\) = \(\left| \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right| \)\([\vec { a } ,\vec { b } ,\vec { c } ]\) = \([\vec { a } ,\vec { b } ,\vec { c } ]\)

Let \(\vec { { F }_{ 1 } } \) and \(\vec { { F }_{ 2 } } \) be the two forces given
Given \(|\vec { { F }_{ 1 } } |=5\sqrt { 2 } \) and its direction is along \(3\hat { i } +4\hat { j } +5\hat { k } \)
∴ \(\vec { { F }_{ 1 } } =5\sqrt { 2 } \) (unit vector of \(3\hat { i } +4\hat { j } +5\hat { k } \))
= \(5\sqrt { 2 } \frac { (3\hat { i } +4\hat { j } +5\hat { k } ) }{ \sqrt { { 3 }^{ 2 }+{ 4 }^{ 2 }+{ 5 }^{ 2 } } } \)   \(\left[ \because \hat { n } =\frac { \vec { n } }{ |\vec { n } | } \right] \)
= \(5\sqrt { 2 } \frac { (3\hat { i } +4\hat { j } +5\hat { k } ) }{ \sqrt { 9+16+25 } } =\frac { 5\sqrt { 2 } (3\hat { i } +4\hat { j } +5\hat { k } ) }{ 5\sqrt { 2 } } \)
= \(3\hat { i } +4\hat { j } +5\hat { k } \)
and  \(\vec { { F }_{ 2 } } =10\sqrt { 2 } \) (unit vector of \(10\hat { i } +6\hat { j } -8\hat { k } \))
= \(10\sqrt { 2 } \frac { (10\hat { i } +6\hat { j } -8\hat { k } ) }{ \sqrt { { 10 }^{ 2 }+{ 6 }^{ 2 }+(-8)^{ 2 } } } \)
\(=\frac{10 \sqrt{\not 2}(10 \hat{i}+6 \hat{j}-8 \hat{k})}{10 \sqrt\not {2}}\)
= \(10\hat { i } +6\hat { j } -8\hat { k } \)
∴ Resistant force \(\vec { F } =\vec { { F }_{ 1 } } +\vec { { F }_{ 2 } } \)
=\((3\hat { i } +4\hat { j } +5\hat { k } )+(10\hat { i } +6\hat { j } -8\hat { k } )\)
\(\vec { F } =13\hat { i } +10\hat { j } -3\hat { k } \)
\(\hat { d } \) = displacement to the point - displacement from the point
= \((6\hat { i } +\hat { j } -3\hat { k } )-(4\hat { i } -3\hat { j } -2\hat { k } )\)
= \(2\hat { i } +4\hat { j } -\hat { k } \)
∴ Work done
w = \(\vec { F } .\vec { d } =(13\hat { i } +10\hat { j } -3\hat { k } ).(2\hat { i } +4\hat { j } -\hat { k } )\)
w = 13(2) + 10(4) - 3(-1) = 26 + 40 + 3
w = 69 units.

Let A be the point (2, 0, -1). Then the position vector of A is  \(\vec { OA } \) = \(2\hat { i } -\hat { k } \) and therefore \(\vec { r } =\vec { OA } =-2\hat { i } -\hat { k } \)

Then the given force is \(\vec{F}=\hat { 2i } +\hat { j } -\hat { k } \) So, the torque is 
\(\vec { t } =\vec { r } \times \vec { F } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 0 & 1 \\ 2 & 1 & -1 \end{matrix} \right| =-\hat { i } -2\hat { k } \)
The magnitude of the torque = \(\left| -\hat { i } -\hat { 2k } \right| =\sqrt { 5 } \) and the direction cosines of the torque are \(-\frac { 1 }{ \sqrt { 5 } } \), 0 , \(-\frac { 2 }{ \sqrt { 5 } } \)

With usual notations in triangle, ABC let \(\vec { BC } =\vec { a } \), \(\vec { CA } =\vec { b } \) and \(\vec { AB } =\vec { c} \). Then \(|\vec { BC }| =\vec { a } \), \(|\vec { CA } |=\vec { b } \) and \(|\vec { AB }| =\vec { c } \)
Since in ΔABC, \(\vec { BC }+\vec { CA }+\vec { AB}=0\) we have \(\vec { BC }\times(\vec { BC }+\vec { CA }+\vec { AB })=\vec { 0 }\)
Simplyfying, we get,
\(\vec { BC }\times\vec { CA }=\vec { AB }\times\vec { BC }\) ....(1)

Similarly, since \(\vec { BC }+\vec { CA }+\vec { AB}=\vec 0\), we have
\(\vec {CA} \times (\vec { BC }+\vec { CA }+\vec { AB})=\vec 0\) .... (2)
On Simplification, we obtain \(\vec { BC }\times\vec { CA }=\vec {CA }\times\vec {AB }\)
From equations (1) and (2), we get 
\(\vec { AB }\times\vec {BC }\) = \(\vec { CA }\times\vec {AB }\)=\(\vec { BC}\times\vec {CA}\)
So, \(\left| \overrightarrow { AB} \times \overrightarrow { BC } \right| =\left| \overrightarrow { CA } \times \overrightarrow { AB } \right| =\left| \overrightarrow { BC } \times \overrightarrow { CA } \right| \). Then, we get
ca sin(π − B) = bc sin(π - A) = ab sin (π - C)
That is, ca sin B = bc sin A = absinC . Dividing by abc, we get
\(\frac { sinA }{ A } =\frac { sinB }{ b } =\frac { sinC }{ c } \) or \(\frac { a }{ sinA }= \frac { b }{ sinB } =\frac { c }{ sinC } \)

With usual notations in triangle ABC, let \(\vec { BC } =\vec { a } ,\vec { CA } =\vec { b } \)  and  \(\vec { AB } =\vec { c } \).
Then \(\left| \vec { BC } \right| =a\) , \(\left| \vec { CA } \right| =b\), \(\left| \vec { AB} \right| =c\), and \(\vec { BC } +\vec { CA } +\vec { AB } =\vec { 0 } \)
So, \(\vec { BC } =-\vec { CA } -\vec { AB } \)
Applying dot product, we get
\(\vec { BC } .\vec { BC } =-\vec { BC } .\vec { CA }-\vec { BC }. \vec { AB } \)
⇒ \({ \left| \vec { BC } \right| }^{ 2 }=-\left| \vec { BC } \right| \left| \vec { CA } \right| \) cos(兀-c)-\(\left| \vec { BC } \right| \left| \vec { AB} \right| \)cos(兀-B)
⇒ a² = ab cos C + ac cos B
Therefore a = b cos C + c cos B
The results (ii) and (iii) are proved in a similar way

Here (x₁, y₁, z₁) = (3, -4, -3) and direction ratios of the given straight line are (a, b, c) = (-4, -7, 12). 
Direction ratios of the normal to the given plane are (A, B, C) = (5, -1, 1).
We observe that, the given point (x₁, y₁, z₁) = (3, 4, -3) satisfies the given plane 5x-y+z = 8
Next, aA+bB+cC = (-4)(5)+(-7)(-1)+(12)(1) = -1 \(\neq \) 0.
So, the normal to the plane is not perpendicular to the line.
Hence, the given line does not lie in the plane.

Given \(\vec { a } =\hat { i } +\hat { j } +\hat { k } \), \(\vec { b } =\hat { i } \), \(\vec { c } ={ c }_{ 1 }\hat { i } +{ c }_{ 2 }\hat { j } +{ c }_{ 3 }\hat { k } \)
∴ \(\vec { c } =\hat { i } +2\hat { j } +{ c }_{ 3 }\hat { k } \)
Also, it given that \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar.
∴ \(\vec { a } .(\vec { b } \times \vec { c } )\)
⇒ \(\left| \begin{matrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & { c }_{ 3 } \end{matrix} \right| \)= 0
⇒ \(1\left| \begin{matrix} 0 & 0 \\ 2 & { c }_{ 3 } \end{matrix} \right| -1\left| \begin{matrix} 1 & 0 \\ 1 & { c }_{ 3 } \end{matrix} \right| +1\left| \begin{matrix} 1 & 0 \\ 1 & 2 \end{matrix} \right| \) = 0
⇒ 1(0-0)-1(c₃-0)+1(2-0) = 0
⇒ 0 - c₃+2 = 0 ⇒ c₃ = 2
∴ c₃ = 2

Let \(\vec { a } =7\hat { i } +\lambda \hat { j } -3\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -\hat { k } \) and \(\vec { c } =-3\hat { i } +7\hat { j } -5\hat { k } \)
∴ volume of the parallelepiped
= \(\vec { a } .(\vec { b } \times \vec { c } )\)
Given \(\vec { a } .(\vec { b } \times \vec { c } )\) = 90
⇒ \(\left| \begin{matrix} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{matrix} \right| \) = 90
⇒ \(-6\left| \begin{matrix} 2 & -1 \\ 7 & 5 \end{matrix} \right| -\lambda \left| \begin{matrix} 1 & -1 \\ -3 & 5 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ -3 & 7 \end{matrix} \right| \) = 90
⇒ 7(10+7)-λ(5-3)-3(7+6) = 90
⇒ 7(17)-λ(2)-3(13) = 90
⇒ 119-2λ-39 = 90
⇒ 119-39-90 = 2λ
⇒ -10 = 2λ
⇒ λ = -5

Here, \(\vec { a } =\hat { i } +\hat { 2j } -\hat { 3k } \), \(\vec { b } =\hat { 2i } -\hat { j } +\hat { 2k } \), \(\vec { c } =\hat { 3i } +\hat { j } -\hat { k } \)
We know that  \(\vec { a } ,\vec { b } ,\vec { c } \)  are coplanar if and only if \([\vec { a } ,\vec { b } ,\vec { c } ]\) = 0. Now, \([\vec { a } ,\vec { b } ,\vec { c } ]\) = \(\left| \begin{matrix} 1 & 2 & -3 \\ 2 & -1 & 2 \\ 3 & 1 & -1 \end{matrix} \right| =0\)
Therefore, the three given vectors are coplanar.

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)

Comparing the given equation with equation of a straight line \(\vec { r } =\vec { a } +t\vec { b } \), we have \(\vec { a } =3\hat { i } -2\hat { j } +6\hat { k } \) and \(\vec { b } =3\hat { i } -2\hat { j } +6\hat { k } \). Therefore,
(i) If \(\vec { b } ={ b }_{ 1 }\hat { i } +{ b }_{ 2 }\hat { j } +{ b }_{ 3 }\hat { k } \), then direction ratios of the straight line are \({ b }_{ 1 },{ b }_{ 2 },{ b }_{ 3 }\). Therefore, direction ratios of the given straight line are proportional to 2, −1,3, and hence the direction cosines of the given straight line are \(\frac { 2 }{ \sqrt { 14 } } ,\frac { -1 }{ \sqrt { 14 } } ,\frac { 3 }{ \sqrt { 14 } } \)
(ii) vector equation of the straight line in non-parametric form is given by \((\vec { r } -\vec { a } )\times \vec { b } =\vec { 0 } \), Therefore, \((\vec{r}-(3\hat { i } -2\hat { j } +6\hat { k } )\times(2\hat { i } -\hat { j } +3\hat { k } )=\vec{0}\)
(iii) Here (x₁, y₁ ,z₁ ) = (3, −2,6) and the direction ratios are proportional to 2, −1,3. Therefore, Cartesian equations of the straight line are \(\frac { x-3 }{ 2 } =\frac { y+2 }{ -1 } =\frac { z-6 }{ 3 } \)

Given \(\vec { a } =2\hat { i } +3\hat { j } -\hat { k } ,\vec { b } =-\hat { i } +2\hat { j } -4\hat { k } ,\vec { c } =\hat { i } +\hat { j } +\hat { k } \)
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ -1 & 2 & -4 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 3 & -1 \\ 2 & -4 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ -1 & -4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ -1 & 2 \end{matrix} \right| \)
= \(\hat { i } (-12+2)-\hat { j } (-8-1)+\hat { k } (4+3)\)
= \(-10\hat { i } +9\hat { j } +7\hat { k } \)
\(\vec { a } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & -1 \\ 1 & 1 & 1 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -1 \\ 1 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -1 \\ 1 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 3 \\ 1 & 1 \end{matrix} \right| \)
= \(\hat { i } (3+1)-\hat { j } (2+1)+\hat { k } (2-3)\)
= \(4\hat { i } -3\hat { j } -\hat { k } \)
∴ \((\vec { a } \times \vec { b } ).(\vec { a } \times \vec { c } )=(-10\hat { i } +9\hat { j } +7\hat { k } ).(4\hat { i } -3\hat { j } -\hat { k } )\)
= -40-27-7 = -74

By definition,
\(\vec { a } \times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 0 \\ 1 & -1 & -4 \end{matrix} \right| =4\hat { i } +4\hat { j } ,\vec { c } \times \vec { d } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 0 & 3 & -1 \\ 2 & 5 & 1 \end{matrix} \right| =8\hat { i } -2\hat { j } -6\hat { k } \)
\((\vec { a } \times \vec { b } )(\vec { c } \times \vec { d } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 4 & 4 & 0 \\ 8 & -2 & -6 \end{matrix} \right| =-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(1)
On the other hand, we have
\([\vec { a }, \vec { b }, \vec { d } ]\vec { c } -[\vec { a } ,\vec { b } ,\vec { c } ]\vec { d } =28(3\vec { j } -\vec { k } )-12(2\hat { i } +5\hat { j } +\hat { k } )=-24\hat { i } +24\hat { j } -40\hat { k } \) ..............(2)
Therefore, from equations (1) and (2), identity (i) is verified.
The verification of identity (ii) is left as an exercise to the reader

(1) \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ d } =p\)
 

(4) Skew lines

(4) velocity