Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

MEDIUM

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IMPORTANT

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Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b},$ where $\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ .

Important Points to Remember in Chapter -1 - Vector Algebra from NCERT MATHEMATICS PART II Textbook for Class XII Solutions

1. Vectors as the sides of a triangle:

If $\vec{a}, \vec{b}, \vec{c}$ are the vectors represented by the sides of a triangle taken in order, then $\vec{a} + \vec{b} + \vec{c} = \vec{0}$ . Conversely, if $\vec{a}, \vec{b}, \vec{c}$ are three non-collinear vectors, such that $\vec{a} + \vec{b} + \vec{c} = \vec{0},$ then they form the sides of a triangle taken in order.

2. Types of vectors:

(i) Co-linear vectors:

Two non-zero vectors $\vec{a}$ and $\vec{b}$ are collinear if there exist non-zero scalars $x$ and $y$ such that $x \vec{a} + y \vec{b} = \vec{0}$ .

(ii) Non-colinear vectors:

If $\vec{a}$ and $\vec{b}$ are two non-zero non-collinear vectors, then $x \vec{a} + y \vec{b} = \vec{0} \Rightarrow$ $x = y = 0 .$

(iii) Co-planar vectors:

If $\vec{a}$ and $\vec{b}$ are two non-zero vectors, then any vector $\vec{r}$ coplanar with $\vec{a}$ and $\vec{b}$ can be uniquely expressed as $\vec{r} = x \vec{a} + y \vec{b},$ where $x, y$ are scalars.

Also, $\vec{r} = (x | \vec{a} |) \hat{a} + (y | \vec{b} |) \hat{b}$

(iv) Representation of a vector $\vec{r}$ in terms of three given non-coplanar vectors:

If $= \pm \frac{(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})}{| (\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) |} = \pm \frac{16 \hat{i} - 16 \hat{j} - 8 \hat{k}}{24}$ are three given non-coplanar vectors, then every vector $\vec{r}$ in space can be uniquely expressed as $\vec{r} = x \vec{a} + y \vec{b} + z \vec{c}$ for some scalars $x, y$ and $z$ .

3. Linearly dependent and independent vectors:

(i) A set of non-zero vectors $\vec{a_{1}}, \vec{a_{2}}, \vec{a_{3}}, \dots, \vec{a_{n}}$ are linearly independent, if $x_{1} \vec{a_{1}} + x_{2} \vec{a_{2}} + \dots . + x_{n} \vec{a_{n}} = 0 \Rightarrow x_{1} = x_{2} = . . . . . . = x_{n} = 0$

(ii) A set of vectors $\vec{a_{1}}, \vec{a_{2}}, \vec{a_{3}}, \dots, \vec{a_{n}}$ are linearly dependent, if there exist scalars $x_{1}, x_{2}, \dots, x_{n}$ not all zero such that $x_{1} \vec{a_{1}} + x_{2} \vec{a_{2}} + \dots . + x_{n} \vec{a_{n}} = 0 .$

(iii) Any two non-zero, non-collinear vectors are linearly independent.

(iv) Any two collinear vectors are linearly dependent.

(v) Any three non-coplanar vectors are linearly independent.

(vi) Any three coplanar vectors are linearly dependent.

4. Section formula:

(i) If $A$ and $B$ are two points with position vectors $\vec{a}$ and $\vec{b}$ respectively, then the position vector of a point $C$ dividing $A B$ in the ratio $m : n$ internally and externally are $\frac{m \vec{b} + n \vec{a}}{m + n}$ and $\frac{m \vec{b} - n \vec{a}}{m - n}$ respectively.

(ii) If $A$ and $B$ are two points with position vectors $\vec{a} and \vec{b}$ respectively and $m, n$ are positive real numbers, then $m \vec{O A} + n \vec{O B} = (m + n) \vec{O C}$ , where $C$ is a point on $A B$ dividing it in the ratio $n : m .$

(iii) Centroid of a triangle:

If $S$ is any point in the plane of a triangle $A B C,$ then $\vec{S A} + \vec{S B} + \vec{S C} = 3 \vec{S G}$ , where $G$ is the centroid of $∆ A B C$

5. Condition for collinearity:

The necessary and sufficient condition for three points with position vectors $\vec{a}, \vec{b}, \vec{c}$ to be collinear is that there exist scalars $x, y, z$ not all zero such that $x \vec{a} + y \vec{b} + z \vec{c} = \vec{0},$ where $x + y + z = 0 .$

6. Condition for points to be co-planar:

The necessary and sufficient condition for four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ to be coplanar is that there exist scalars $x, y, z, t$ not all zero such that $x \vec{a} + y \vec{b} + z \vec{c} + t \vec{d} = \vec{0},$ where $x + y + z + t = 0$ .

7. Dot product of two vectors:

If $\vec{a}$ and $\vec{b}$ are two non-zero vectors inclined at an angle $θ,$ then

(i) $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos θ$

(ii) Projection of $\vec{a}$ on $\vec{b} = \frac{\vec{a} \cdot \vec{b}}{| \vec{b} |} = \vec{a} \cdot \hat{b}$

(iii) Projection of $\vec{b}$ on $\vec{a} = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} |} = \vec{b} \cdot \hat{a}$

(iv) Projection vector of $\vec{a}$ on $\vec{b}$ $= \{\frac{\vec{a} \cdot \vec{b}}{| \vec{b} |}\} \hat{b} = \{\frac{\vec{a} \cdot \vec{b}}{| \vec{b} |^{2}}\} \vec{b}$

(v) Projection vector of $\vec{b}$ on $\vec{a}$ $= \{\frac{\vec{a} \cdot \vec{b}}{| \vec{a} |}\} \hat{a} = \{\frac{\vec{a} \cdot \vec{b}}{| \vec{a} |^{2}}\} \vec{a}$

(vi) $\vec{a} \cdot \vec{b} = 0 \Leftrightarrow \vec{a}$ is perpendicular to $\vec{b}$

(vii) $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$

(viii) $\vec{a} \cdot \vec{a} = | \vec{a} |^{2}$

(ix) $m \vec{a} \cdot \vec{b} = m (\vec{a} \cdot \vec{b}) = \vec{a} \cdot m \vec{b}$ , for any scalar $m$

(x) $m \vec{a} \cdot n \vec{b} = m n (\vec{a} \cdot \vec{b}) = m n (\vec{a} \cdot \vec{b}) = \vec{a} \cdot m n \vec{b}$ for scalars $m, n$

(xi) $| \vec{a} \pm \vec{b} | \leq | \vec{a} | + | \vec{b} |$

(xii) $| \vec{a} - \vec{b} | \geq | \vec{a} | - | \vec{b} |$

(xiii) $| \vec{a} \pm \vec{b} |^{2} = | \vec{a} |^{2} + | \vec{b} |^{2} \pm 2 (\vec{a} \cdot \vec{b})$

(xiv) $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = | \vec{a} |^{2} - | \vec{b} |^{2}$

(xv) $\vec{a} \cdot \vec{b} > 0$ if $θ$ is acute

(xvi) $\vec{a} \cdot \vec{b} < 0$ if $θ$ is obtuse

(xvii) Dot product in component form:

If $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ , then $\vec{a} \cdot \vec{b}$ $= a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3} .$

8. Angle between two vectors:

If $\vec{a}$ and $\vec{b}$ are two vectors inclined at an angle $θ,$ then $c o s θ = \frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}$ .

9. Scalar Triple Product:

(i) Properties of scalar triple product:

(a) The dot Product of the vector $\vec{a}$ $\times$ $\vec{b}$ with the vector $\vec{c}$ is scalar triple product of three vectors $\vec{a}$ , $\vec{b}$ , $\vec{c}$ and it is written as $(\vec{a} \times \vec{b}) \cdot \vec{c}$ it is a scalar quantity.

(b) $(\vec{a} \times \vec{b}) \cdot \vec{c} = \vec{a} \cdot (\vec{b} \times \vec{c})$ i.e. dot and cross can be interchanged in a scalar triple product is written as $[\vec{a} \vec{b} \vec{c}]$ .

(c) $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}] = - [\vec{b} \vec{a} \vec{c}] = - [\vec{c} \vec{b} \vec{a}] = - [\vec{a} \vec{c} \vec{b}]$

(d) If $\vec{a} = (a_{1}, a_{2}, a_{3}) = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}, \vec{b} = (b_{1}, b_{2}, b_{3}) = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$ and $\vec{c} = (c_{1}, c_{2}, c_{3}) = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k},$ then $[\vec{a} \vec{b} \vec{c}] = |\begin{matrix} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{matrix}|$

(ii) The volume of the Parallelepiped:

(a) with $\vec{a}, \vec{b}, \vec{c}$ as Coterminous edges = $|[\vec{a} \vec{b} \vec{c}]|$

(b) with $A, B, C, D$ as vertices of coterminous edges is $|[\vec{A B} \vec{A C} \vec{A D}]|$ cubic units.

10. Cross product of two vectors:

(i) If $\vec{a}, \vec{b}$ are two vectors inclined at an angle $θ$ , then $\vec{a} \times \vec{b} = | \vec{a} | | \vec{b} | \sin θ \hat{n}$ , where $\hat{n}$ is a unit vector perpendicular to the plane of $\vec{a}$ and $\vec{b}$ such that $\vec{a}, \vec{b}, \hat{n}$ form a right handed system.

(ii) $\vec{a} \times \vec{b} = \vec{0}$ if $\vec{a}$ and $\vec{b}$ are parallel.

(iii) Area of a parallelogram:

$\vec{a} \times \vec{b}$ gives the vector area of the parallelogram having two adjacent sides as $\vec{a}$ and $\vec{b}$ .

(iv) For any two co-planar vectors $\vec{a} and \vec{b}$ , $\vec{a} \times \vec{b} = - (\vec{b} \times \vec{a})$

(vi) $m \vec{a} \times n \vec{b} = m n (\vec{a} \times \vec{b}) = n \vec{a} \times m \vec{b} = m n \vec{a} \times \vec{b} = \vec{a} \times m n \vec{b}$ for scalars $m, n .$

11. Cross product in component form:

$\vec{a} \times \vec{b} = |\begin{array}{l} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}|$ , where $\vec{a} = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ and $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

12. Area of plane figures:

(i) Area of a triangle:

Area of $∆ A B C$ $= \frac{1}{2} |\vec{A B} \times \vec{A C}| = \frac{1}{2} |\vec{B C} \times \vec{B A}| = \frac{1}{2} |\vec{C B} \times \vec{C A}|$

(ii) Area of a quadrilateral:

Area of a plane convex quadrilateral $A B C D$ is $\frac{1}{2} |\vec{A C} \times \vec{B D}|$ , where $A C$ and $B D$ are diagonal.

(iii) Area of a triangle when the vertices are given:

If $\vec{a}, \vec{b}, \vec{c}$ are the position vectors of the vertices $A, B, C$ of $∆ A B C,$ then

Area of $∆ A B C = \frac{1}{2} |\vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a}|$

Find a unit vector perpendicular to each of the vector a→+b→ and a→-b→, where a→=3i^+2j^+2k^ and b→=i^+2j^-2k^.

Important Points to Remember in Chapter -1 - Vector Algebra from NCERT MATHEMATICS PART II Textbook for Class XII Solutions

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Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b},$ where $\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ .