Concept:

Unit vector parallel to $\overrightarrow{a}=\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

Calculation:

Let $\overrightarrow{a}$ = -2î + 3ĵ

$\Rightarrow \left|\overrightarrow{a}\right|=\sqrt{(-2{)}^2+3^2}=\sqrt{13}$

∴ Unit vector parallel to $\overrightarrow{a}=\hat{a}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}$

$\Rightarrow \frac{1}{\sqrt{13}}(-2\hat{i}+3\hat{j})$

$\Rightarrow \frac{-2\hat{i}}{\sqrt{13}}+\frac{3\hat{j}}{\sqrt{13}}$

Concept:

Direction cosine:

• In 3-D geometry, there are three axes, namely x-axes, y-axes and z-axes.  
• If line OP passes through the origin and makes angle α, β and γ from x, y and z axes respectively.
• The cosines of each of these angles that the line makes with the x-axis, y-axis, and z-axis respectively are called direction cosines.  

​Therefore, cosα, cosβ and cosγ are direction cosine and

⇒ cos2α + cos2β + cos2γ = 1

Calculation:

Let direction cosines are (cosα, cosβ, cosγ)

We know that,

cos2α + cos2β + cos2γ = 1    ----(1)

According to the question, the angles are equal,

α = β = γ = θ

Hence, from equation (1)

cos2θ + cos2θ + cos2θ = 1

⇒ 3cos2θ = 1

⇒ $cos\theta =\pm \frac{1}{\sqrt{3}}$

Hence, direction cosines are

(cosα, cosβ, cosγ) = $\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}},\pm \frac{1}{\sqrt{3}}$

Concept:

• The shortest distance between the lines $\overrightarrow{r}={\overrightarrow{A}}_1+\lambda {\overrightarrow{B}}_1$  and $\overrightarrow{r}={\overrightarrow{A}}_2+\lambda {\overrightarrow{B}}_2$ is given by,

$d=|({\overrightarrow{A}}_2-{\overrightarrow{A}}_1).\widehat{({\overrightarrow{B}}_1\times {\overrightarrow{B}}_2)}|$

$\Rightarrow d=|\frac{({\overrightarrow{A}}_2-{\overrightarrow{A}}_1).({\overrightarrow{B}}_1\times {\overrightarrow{B}}_2)}{|{\overrightarrow{B}}_1\times {\overrightarrow{B}}_2|}|$

• Where $\widehat{({\overrightarrow{B}}_1\times {\overrightarrow{B}}_2)}$ is the unit vector perpendicular to ${\overrightarrow{B}}_{1}and{\overrightarrow{B}}_2$
• Here.  ${\overrightarrow{B}}_{1}and{\overrightarrow{B}}_2$ are representing the direction of the first and second lines ${\overrightarrow{A}}_{1}and{\overrightarrow{A}}_2$ are representing the points on the first and second lines

The above can also be written in the form,

• If the two lines are intersecting then,

d = 0

Calculation:

• For intersecting lines,

$d=|\frac{({\overrightarrow{A}}_2-{\overrightarrow{A}}_1).({\overrightarrow{B}}_1\times {\overrightarrow{B}}_2)}{|{\overrightarrow{B}}_1\times {\overrightarrow{B}}_2|}|=0$

$\Rightarrow |({\overrightarrow{A}}_2-{\overrightarrow{A}}_1).({\overrightarrow{B}}_1\times {\overrightarrow{B}}_2)|=0$

• Hence, option 4 is correct.

Concept:

Projection of any vector  $\overrightarrow{\mathrm{x}}$ on vector $\overrightarrow{\mathrm{y}}$ is:

P = $\overrightarrow{\mathrm{x}}\cdot \hat{\mathrm{y}}$

Where $\hat{\mathrm{y}}$ is unit vector in direction of vector $\overrightarrow{\mathrm{y}}$

$\hat{\mathrm{y}}$ = $\frac{\overrightarrow{\mathrm{y}}}{|\overrightarrow{\mathrm{y}}|}$

Calculation:

Given $\overrightarrow{\mathrm{a}}$ = i + 2j + k and $\overrightarrow{\mathrm{b}}$ = -i + j - 3k

Projection of $\overrightarrow{\mathrm{a}}$ on $\overrightarrow{\mathrm{b}}$ (let P)= $\overrightarrow{\mathrm{a}}\cdot \hat{\mathrm{b}}$

Now, unit vector in direction of vector $\overrightarrow{\mathrm{b}}$ is $\hat{\mathrm{b}}=\frac{\overrightarrow{\mathrm{b}}}{|\overrightarrow{\mathrm{b}}|}$

$\hat{\mathrm{b}}=\frac{-\mathrm{i}+\mathrm{j}-3\mathrm{k}}{\sqrt{(-1{)}^2+1^2+(-3{)}^2}}$

$\hat{\mathrm{b}}=\frac{1}{\sqrt{11}}$(-i + j - 3k)

Now P = (i + 2j + k) ⋅ $\frac{1}{\sqrt{11}}$(-i + j - 3k)

P = $\frac{1}{\sqrt{11}}$(-1 + 2 - 3)

P = $\frac{\mathbf{-}\mathbf{2}}{\sqrt{\mathbf{11}}}$

Concept:

• If a, b, c are coplanar then [a b c] = 0
• Three vectors are permuted in the same cyclic order, the value of the scalar triple product remains the same. ⇒  [a b c] = [b c a] = [c a b]

Calculation:

Here, a, b, c are non-coplanar

To find: 2[a b c] + [b a c] =?

⇒ 2[a b c] + [b a c]

= 2 [a, b, c] - [a, b, c]                (∵ [b a c] = -[a b c])

= [a, b, c] 

Hence, option (2) is correct.

Concept:

Equation of a line with direction ratio (a, b, c) that passes through the point (x₁, y₁, z₁) is given by the formula: $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

Equation of plane in 3-D: lx + my + nz = d

Direction ratios of normal: l, m, n

If two Vectors are perpendicular then the dot product of their direction ratios are equal to zero.

Calculation:

Here, equation of plane ax + by + cz + d = 0,

Direction ratios of the normal = a, b, c

Equation of the straight line $\frac{x-x_0}{l}=\frac{y-y_0}{m}=\frac{z-z_0}{n}$

Direction ratios = l, m, n

Since, The line and plane are parallel to each other 

So, The line and Normal of the plane are perpendicular to each other 

So, the dot product of their direction ratios = 0

∴ al + bm + cn = 0

Concept:

• ${\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2=2\times \left({\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2\right)$

Calculation:

Given:  $|\overrightarrow{a}|$ = 3, $\left|\overrightarrow{b}\right|=4$ and $\left|\overrightarrow{a}-\overrightarrow{b}\right|=5,$

We know that,

${\left|\overrightarrow{a}-\overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2=2\times \left({\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2\right)$

$\Rightarrow 5^2+{\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2=2\times \left(3^2+4^2\right)$

$\Rightarrow {\left|\overrightarrow{a}+\overrightarrow{b}\right|}^2=50-25=25$

∴ $\left|\overrightarrow{a}+\overrightarrow{b}\right|=5$

Concept:

Let us consider two lines AB and CD. The direction ratios of line AB is a1, b1, c1 and the direction ratios of line CD is a2, b2, c2.

Then AB will be parallel to CD, if  $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$.

Calculation:

Given: The line through the points (2, 4, 8) and (1, 2, 4) is parallel to the line through the points (3, 6, k) and (1, 2, 1).

Let us consider AB be the line joining the points (2, 4, 8) and (1, 2, 4) whereas CD be the line passing through the points (3, 6, k) and (1, 2, 1).

Let, the direction ratios of AB be: a1, b1, c1 

⇒ a₁ = (2 – 1) = 1, b₁ = (4 – 2) = 2 and c₁ = (8 – 4) = 4.

Let the direction ratios of CD be: a2, b2, c2 

⇒ a₂ = (3 – 1) = 2, b₂ = (6 – 2) = 4 and c₂ = k – 1.

∵ Line AB is parallel to CD ⇒  $\frac{{\mathrm{a}}_1}{{\mathrm{a}}_2}=\frac{{\mathrm{b}}_1}{{\mathrm{b}}_2}=\frac{{\mathrm{c}}_1}{{\mathrm{c}}_2}$

⇒ $\frac{1}{2}=\frac{2}{4}=\frac{4}{\mathrm{k}-1}$

⇒ $\frac{1}{2}=\frac{4}{\mathrm{k}-1}$

⇒ k - 1 = 8 ⇒ K = 9.

Hence, correct option is 2.

Concept:

Conditions of collinear vector:

• Three points with position vectors $\overrightarrow{a},\overrightarrow{b}and\overrightarrow{c}$ are collinear if and only if the vectors $\left(\overrightarrow{a}-\overrightarrow{b}\right)$ and $\left(\overrightarrow{a}-\overrightarrow{c}\right)$ are parallel. ⇔ $\left(\overrightarrow{a}-\overrightarrow{b}\right)=\lambda \left(\overrightarrow{a}-\overrightarrow{c}\right)$
• If the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) be collinear then $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

Solution:

We know that, If the points (x1, y1, z1), (x2, y2, z2), and (x3, y3, z3) be collinear then $\left|\begin{array}{ccc}{x}_1& y_1& z_1\\ x_2& y_2& z_2\\ x_3& y_3& z_3\end{array}\right|=0$

Given  $\hat{i}+2\hat{j}+3\hat{k}$, $\lambda \hat{i}+4\hat{j}+7\hat{k}$, $-3\hat{i}-2\hat{j}-5\hat{k}$ are collinear

∴ $\left|\begin{array}{ccc}1& 2& 3\\ \lambda & 4& 7\\ -3& -2& -5\end{array}\right|=0$

⇒ 1 (-20 + 14) – (2) (-5λ + 21) + 3 (-2λ + 12) = 0

⇒ -6 + 10λ – 42 - 6λ + 36  = 0

⇒ 4λ = 12

∴ λ = 3

Concept:

Let P and Q be the given two points (x1, y1) and (x2, y2) respectively, and M be the point dividing the line segment PQ internally in the ratio m : n, then from the section formula, the coordinate of the point M is given by:

$M(x,y)=\{\left(\frac{m{x}_2+n{x}_1}{m+n}\right),\left(\frac{m{y}_2+n{y}_1}{m+n}\right)\}$

Let point P be the point that lies at the y-axis and divide the line segment made by two points A and B in the ratio k : 1.

Since point P lies on the y-axis, therefore, the coordinates of the point P would be of the form (0, y).

Now, using the section formula and equating the x-coordinates, we get

$0=\frac{3k-2}{k+1}$

⇒ 3k - 2 = 0

⇒ k = 2/3

∴ k : 1 = 2 : 3

Hence, the required ratio is 2 : 3.