Applications of Vector Algebra Model Question Paper

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Questions:

Generic 3

3 x 3 = 9

If \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\) then show that \(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)

If \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } ,\overset { \rightarrow }{ b } =\overset { \wedge }{ j } -\overset { \wedge }{ k } ,\overset { \rightarrow }{ c } =\overset { \wedge }{ k } -\overset { \wedge }{ i } \) then find \(\left[ \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ b } -\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right] \)

With usual notations, in any triangle ABC, prove the following by vector method.
(i) a = b cos C + c cos B
(ii) b = c cos A + a cos C
(iii) c = a cos B + b cos A

Generic 5

3 x 5 = 15

Find the vector and Cartesian equation of the plane passing through the point (1,1, -1) and perpendicular to the planes x + 2y + 3z - 7 = 0 and 2x - 3y + 4z = 0

Find the distance of the point (5, -5, -10) from the point of intersection of a straight line passing through the points A (4, 1, 2) and B (7, 5, 4) with the plane x - y + z = 5

In triangle, ABC the points, D, E, F are the midpoints of the sides BC, CA and AB respectively. Using vector method, show that the area of ΔDEF is equal to \(\frac{1}{4}\)(area of ΔABC )

Generic 2

4 x 2 = 8

Find the equation of the plane containing the line of intersection of the planes x + y + Z - 6 = 0 and 2x + 3y + 4z + 5 = 0 and passing through the point (1, 1, 1)

The volume of the parallelepiped whose coterminus edges are \(7\hat { i } +\lambda \hat { j } -3\hat { k } ,\hat { i } +2\hat { j } -\hat { k } \), \(-3\hat { i } +7\hat { j } +5\hat { k } \) is 90 cubic units. Find the value of λ.

If \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \) find \(\vec { a } .(\vec { b } \times \vec { c } )\).

10.

If \(\hat { 2i } -\hat { j } +\hat { 3k } ,\hat { 3i } +\hat { 2j } +\hat { k } ,\hat { i } +\hat { mj } +\hat { 4k } \) are coplanar, find the value of m.

Multiple Choice Question

5 x 1 = 5

11.

The distance from the origin to the plane \(\overset { \rightarrow }{ r } .\left( \overset { \wedge }{ 2i } -\overset { \wedge }{ j } +5\overset { \wedge }{ k } \right) =7\) is ______________

(a)

\(\frac { 7 }{ \sqrt { 30 } } \)

(b)

\(\frac { \sqrt { 30 } }{ 7 } \)

(c)

\(\frac { 30 }{ 7 } \)

(d)

\(\frac { 7 }{ 30 } \)

12.

The angle between the vector \(3\overset { \wedge }{ i } +4\overset { \wedge }{ j } +\overset { \wedge }{ 5k } \) and the z-axis is ___________

(a)

30^o

(b)

60^o

(c)

45^o

(d)

90^o

13.

The number of vectors of unit length perpendicular to the vectors \(\left( \overset { \wedge }{ i } +\overset { \wedge }{ j } \right) \) and \(\left( \overset { \wedge }{ j } +\overset { \wedge }{ k } \right) \)is __________

(a)

(b)

(c)

(d)

\(\infty\)

14.

If \(\vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } \times \vec { b } )\times \vec { c } \) where \(\vec { a } ,\vec { b } ,\vec { c } \) are any three vectors such that \(\vec{b} \cdot \vec{c} \neq 0 \text { and } \vec{a} \cdot \vec{b} \neq 0\), then \(\vec { a } \) and \(\vec { c } \) are

(a)

perpendicular

(b)

parallel

(c)

inclined at an angle \(\frac{\pi}{3}\)

(d)

inclined at an angle \(\frac{\pi}{6}\)

15.

If \([\vec{a}, \vec{b}, \vec{c}]=1\), then the value of \(\frac{\vec{a} \cdot(\vec{b} \times \vec{c})}{(\vec{c} \times \vec{a}) \cdot \vec{b}}+\frac{\vec{b} \cdot(\vec{c} \times \vec{a})}{(\vec{a} \times \vec{b}) \cdot \vec{c}}+\frac{\vec{c} \cdot(\vec{a} \times \vec{b})}{(\vec{c} \times \vec{b}) \cdot \vec{a}}\) is

(a)

(b)

-1

(c)

(d)

Odd one out

1 x 2 = 2

16.

\(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) are said to be coplanar if
(1) \(\left[ \overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } \right] \)=0
(2) \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } \) lie on the same plane
(3) They are either parallel or intersecting
(4) Skew lines

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Answers Key:

Generic 3

3 x 3 = 9

Given \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\)
\(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \)= -\(\overset { \rightarrow }{ c } \) ... (1)
Taking cross product with \(\overset { \rightarrow }{ a } \) both sides, we get
\(\overset { \rightarrow }{ a } \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =-\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)
\(\left[ \because -\left( \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ c } \right) =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \right] \)
\(\Rightarrow \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \left( \because \overset { \rightarrow }{ a } \times \overset { \rightarrow }{ a } =\overset { \rightarrow }{ 0 } \right) \)
Taking cross product with \(\overset { \rightarrow }{ b } \) both sides, we get
\(\overset { \rightarrow }{ b } \times \left( \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \right) =\overset { \rightarrow }{ b } \times \left( -\overset { \rightarrow }{ c } \right) \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(-\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =-\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \)
From (2) and (3) we get
\(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)

\(\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } \)= \(\overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } \right) -\left( \overset { \wedge }{ j } -\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } +\overset { \wedge }{ k } \right) \)
\(\overset { \rightarrow }{ b } -\overset { \rightarrow }{ c } =\left( \overset { \wedge }{ j } -\overset { \wedge }{ k } \right) -\left( \overset { \wedge }{ k } -\overset { \wedge }{ i } \right) =\overset { \wedge }{ i } +\overset { \wedge }{ j } -2\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =\left( \overset { \wedge }{ k } -\overset { \wedge }{ i } \right) -\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } \right) =-2\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
\(\therefore \left[ \overset { \rightarrow }{ a } -\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ b } -\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right] =\left| \begin{matrix} 1 \\ 1 \\ -2 \end{matrix}\begin{matrix} 0 \\ 1 \\ 1 \end{matrix}\begin{matrix} 1 \\ -2 \\ 1 \end{matrix} \right| \)
= 1 ( 1 + 2) + 0 + 1( 1+ 2)
= 3 + 3 = 6

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

Generic 5

3 x 5 = 15

The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \)
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

The Cartesian equation of the straight line joining A and B is
\(\frac { x-4 }{ 3 } =\frac { y-1 }{ 4 } =\frac { z-2 }{ 2 } \) = t (say)
Therefore, an arbitrary point on the straight line is of the form (3t + 4, 4t + 1, 2t + 2).
To find the point of intersection of the straight line and the plane, we substitute x = 3t + 4, y = 4t + 1, z = 2t + 2 in x -y + z = 5 and we get t = 0 Therefore, the point of intersection of the straight line is (4, 1, 2)
Now, the distance between the two points (4, 1, 2) and (5, -5, -10) is
\(\sqrt { (4-5)^{ 2 }+(1+5)^{ 2 }+(2+10)^{ 2 } } \) = \(\sqrt { 181}\) units.

In triangle ABC, consider A as the origin. Then the position vectors of D, E, F are given by \(\frac { \vec { AB } +\vec { AC } }{ 2 } ,\frac { \vec { AC } }{ 2 } ,\frac { \vec { AB } }{ 2 } \) respectively.
Since \(\left| \vec { AB } \times \vec { AC } \right| \) is the area of the parallelogram formed by the two vectors \(\vec { AB }\), \(\vec { AC } \) as adjacent sides, the area of ΔABC is \(\frac{1}{2}\) \(\left| \vec { AB } \times \vec { AC } \right| \). Similarly, considering ΔDEF, we get

the area of ΔDEF = \(\frac{1}{2}\) \(\left| \vec { DE } \times \vec { DF } \right| \)
= \(\frac{1}{2}\) \(\left| (\vec { AE }-\vec{AD}) \times (\vec { AF }-\vec{AD}) \right|\)
= \(\left| \frac { \vec { AB } }{ 2 } \times \frac { \vec { AC } }{ 2 } \right| \)
= \(\frac14\) \(\left( \frac { 1 }{ 2 } \left| \vec { AB } \times \vec { AC } \right| \right) \)
= \(\frac14\)(the area of ΔABC)

Generic 2

4 x 2 = 8

The equation of the required plane through the intersection of the given planes is
( x + y + z - 6 ) + λ (2x + 3y + 4z + 5) = 0 ......(1)
This passes through (1, 1, 1)
∴ ( 1+ 1 + z - 6) + λ (2 + 3+ 4 + 5) = 0
⇒ -3 +14λ = 0 \(\Rightarrow \lambda =\frac { 3 }{ 14 } \)
Substituting \(\lambda =\frac { 3 }{ 14 } \) in (1) we get
( x + y + z - 6 )+\(\frac { 3 }{ 14 } \) (2x + 3y + 4z + 5) = 0
⇒ 14( x + y + z - 6 ) +3 (2 + 3+ 4 + 5) = 0
⇒ 20x + 23y + 26z - 69 = 0

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

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= 1(1+4)+2(2+6)+3(4-3)
= 1(5)+2(8)+3(1)
= 5+16+3 = 24
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 24

10.

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

Multiple Choice Question

5 x 1 = 5

11.