Maths 3 Marks

1. Equation of the first line is $\vec{\text{r}}=2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}+\lambda\Big(3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}\Big)$

Comparing with $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big),$

$ \vec{\text{a}_1}=2\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\ \text{and}\ \vec{\text{b}_1}=3\hat{\text{i}}+2\hat{\text{j}}+6\hat{\text{k}}$

(Vector $\vec{\text{a}}$ is the position vector of a point on line $\vec{\text{b}}$ is a vector along the line)

Again, equation of the second line is $\vec{\text{r}}=7\hat{\text{i}}-6\hat{\text{k}}+\mu\Big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\Big)$

Comparing with $\Big(\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{b}}\Big),$

$\vec{\text{a}_2}=7\hat{\text{i}}-6\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$

(Vector $\vec{\text{a}}$ is the position vector of a point on line $\vec{\text{b}}$ is a vector along the line)

Let $\theta$ be the angle between these two lines, then

$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{3(1)+2(2)+6(2)}{\sqrt{9+4+36}\sqrt{1+4+4}}=\frac{3+4+12}{\sqrt{49}\sqrt{9}}=\frac{19}{7\times3}$

$\cos\theta=\frac{19}{21}\ \ \ \ \ \ \ \Rightarrow\ \theta=\cos^{-1}\frac{19}{21}$

The line is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+9\hat{\text{k}}+5\hat{\text{k}}.$

Now,

$\therefore\vec{\text{b}}\times\overrightarrow{\text{PQ}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&9&5\\-4&-4&1\end{vmatrix}$

$=29\hat{\text{i}}-22\hat{\text{j}}+28\hat{\text{k}}$

$\Rightarrow\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|=\sqrt{29^2+(-22)^2+28^2}$

$=\sqrt{841+484+784}$

$=\sqrt{2109}$

$\big|\vec{\text{b}}\big|=\sqrt{2^+9^2+5^2}$

$=\sqrt{4+81+25}$

$=\sqrt{110}$

$\text{d}=\frac{\big|\vec{\text{b}}\times\overrightarrow{\text{PQ}}\big|}{\big|\vec{\text{b}}\big|}$

$=\frac{\sqrt{2109}}{\sqrt{110}}$

$\vec{\text{a}}=2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}=(2,-1,\ 4)=(\text{x}_1,\ \text{y}_1,\ \text{z}_1)$

The required line is in the direction of the vector $\hat{\text{b}}=\hat{\text{i}}+2\hat{\text{j}-\hat{\text{k}}}$

$\therefore$ Direction ratios of the required line are coefficients of $\hat{\text{i}},\ \hat{\text{j}},\ \hat{\text{k}}\ \text{in}\ \vec{\text{b}}=1,\ 2,-1=\text{a},\ \text{b},\ \text{c}$

$\therefore$ Equation of the required line in vector form is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$

$\Rightarrow\ \ \vec{\text{r}}=\Big(2\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\Big)$

Where $\lambda$ is a real number.

Cartesian equation of this equation is $\frac{\text{x}-2}{1}=\frac{\text{y}+1}{2}=\frac{\text{z}-4}{1}.$

The direction ratios of the line passing through the points (1, -1, 2) and (3, 4, -2) are (3 - 1), [4 - (-1)], (-2 - 2),

i.e. ⇒ a₁ = 2, b₁ = 2, c₁ = -4

Similarly, the direction ratios of the line passing through the points (0, 3, 2) and (3, 5, 6) and (3 - 0), (5 - 3), (6 - 2),

i.e. ⇒ a₂ = 3, b₂ = 2, c₂ = 4

$\therefore$ a₁a₂ + b₁b₂ + c₁c₂ = 2 × 3 + 5 × 2 (-4) × 4 = 6 + 10 - 16 = 0  

Thus the line through the points (1, -1, 2) and (3, 4, -2) is perpendicular to the line throught the points (0, 3, 2) and (3, 5, 6).

2. Given: Equation of first line is $\frac{\text{x}}{2}=\frac{\text{y}}{2}=\frac{\text{z}}{1}$

The direction ratios of this line i.e., a vector along the line is

$\vec{\text{b}_1}=(2,\ 2,\ 1)=2\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$

Now equation of second line is $\frac{\text{x}-5}{4}=\frac{\text{y}-2}{1}=\frac{\text{z}-3}{8}$

The direction ratios of this line i.e., a vector along the line is

$\vec{\text{b}_2}=(4,\ 1,\ 8)=4\hat{\text{i}}+\hat{\text{j}}+8\hat{\text{k}}$

Let $\theta$ be the angle between these two lines, then

$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{2(4)+(2)(1)+(1)(8)}{\sqrt{4+4+1}\sqrt{16+1+64}}=\frac{8+2+8}{\sqrt{9}\sqrt{81}}=\frac{18}{3\times9}=\frac{2}{3}$

$\Rightarrow\ \ \ \theta=\cos^{-1}\frac{2}{3}$

2x + 3y - z = 6

Dividng both sides by 6, we get

$\frac{2\text{x}}{6}+\frac{3\text{y}}{6}-\frac{\text{z}}{6}=\frac{6}{6}$

$\Rightarrow\frac{\text{x}}{3}+\frac{\text{y}}{2}+\frac{\text{z}}{-6}=1\ ...(\text{i})$

We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$

Comparing (i) and (ii) we get

$\text{a}=3;\text{ b}=2;\text{ c}=-6$

$\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\3-2&-2-1&-2-0\\3-2&1-1&7-0\end{vmatrix}=0$

$\Rightarrow\begin{vmatrix}\text{x}-2&\text{y}-1&\text{z}-0\\1&-3&-2\\1&0&7\end{vmatrix}=0$

$\Rightarrow-21(\text{x}-2)-9(\text{y}-1)+3\text{z}=0$

$\Rightarrow-21\text{x}+42-9\text{y}+9+3\text{z}=0$

$\Rightarrow-21\text{x}-9\text{y}+3\text{z}+51=0$

$\Rightarrow21\text{x}+9\text{y}-3\text{z}=51$

$\Rightarrow7\text{x}+3\text{y}-\text{z}=17$ 

If the line meets the plane, then this point must satisfy the equation of plane for some value of $\lambda$

$\therefore(2\lambda-1)+(3\lambda-2)+4(4\lambda-3)=6\text{ }\text{ }\Rightarrow\lambda=1$

$\therefore$Coordinates of required point are (1, 1, 1).

2. Comparing the first and second equation with $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)\ \text{and}\ \Big(\vec{\text{r}}=\vec{\text{a}}+\mu\vec{\text{b}}\Big)$ resp.

$ \vec{\text{b}_1}=\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=3\hat{\text{i}}-5\hat{\text{j}}-4\hat{\text{k}}$

Let $\theta$ be the angle between these two lines, then

$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{1(3)+(-1)(-5)+(-2)(-4)}{\sqrt{1+1+4}\sqrt{9+25+16}}=\frac{3+5+8}{\sqrt{6}\sqrt{50}}=\frac{16}{\sqrt{300}}$

$\cos\theta=\frac{16}{10\sqrt{3}}=\frac{8}{5\sqrt{3}}\ \ \ \ \ \ \ \Rightarrow\ \theta=\cos^{-1}\frac{8}{5\sqrt{3}}.$

2x - y + z = 5

Dividng both sides by 5, we get

$\frac{2\text{x}}{5}+\frac{-\text{y}}{5}+\frac{\text{z}}{5}=\frac{5}{5}$

$\Rightarrow\frac{\text{x}}{\big(\frac{5}{2}\big)}+\frac{\text{y}}{-5}+\frac{\text{z}}{5}=1\ ...(\text{i})$

We know that the equation of the plane whose intercepts on the coordianate axes are a, b and c is,

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ...(\text{ii})$

Comparing (i) and (ii) we get

$\text{a}=\frac{5}{2};\text{ b}=-5;\text{ c}=5$

1. Given: Equation of first line is $\frac{\text{x}-2}{2}=\frac{\text{y}-1}{5}=\frac{\text{z}+3}{-3}$

The direction ratios of this line i.e., a vector along the line is

$\vec{\text{b}_1}=(2,\ 5,-3)=2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}}$

Now equation of second line is $\frac{\text{x}+2}{-1}=\frac{\text{y}-4}{8}=\frac{\text{z}-5}{4}$

The direction ratios of this line i.e., a vector along the line is

$\vec{\text{b}_2}=(-1,\ 8,\ 4)=-\hat{\text{i}}+8\hat{\text{j}}+4\hat{\text{k}}$

Let $\theta$ be the angle between these two lines, then

$\cos\theta=\frac{\vec{\text{b}_1}.\vec{\text{b}_2}}{\Big|\vec{\text{b}_1}\Big|.\Big|\vec{\text{b}_2}\Big|} =\frac{2(-1)+(5)(8)+(-3)(4)}{\sqrt{4+25+9}\sqrt{1+64+16}}=\frac{-2+40-12}{\sqrt{38}\sqrt{81}}=\frac{26}{9\sqrt{38}}$

$\Rightarrow\ \ \ \theta=\cos^{-1}\frac{26}{9\sqrt{38}}.$

The given equations of the planes are

$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})=5$

$\vec{\text{r}}\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})=5$

$\Rightarrow\vec{\text{n}_1}=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}}),$

$\Rightarrow\vec{\text{n}_2}=(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})$

Now, $\vec{\text{n}_1}\cdot\vec{\text{n}_2}=(2\hat{\text{i}}-\hat{\text{j}}+3\hat{\text{k}})\cdot(2\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})$

$=4+2-6=0$

So, the normals to the given planes are perpendicular to each other. 

The equation of the line AB can be re-written as

$\frac{\text{x}-\frac{1}{2}}{1}=\frac{\text{y}-4}{-7}=\frac{\text{z}+1}{2}$

The direction ratios of the line parallel to AB are proportional to 1, -7, 2.

Also, the diraction cosines of the line parallel to AB are proportional to

$\frac{1}{\sqrt{1^2+(-7)^2+2^2}},\frac{-7}{\sqrt{1^2+(-7)^2+2^2}},\frac{2}{\sqrt{1^2+(-7)^2+2^2}}$

$=\frac{1}{\sqrt{54}},\frac{-7}{\sqrt{54}},\frac{2}{\sqrt{54}}$

$\frac{\text{x}-2}{2}=\frac{2\text{y}-5}{-3},\text{z}=2$

The equation of given line can be re-written as

$\frac{\text{x}-2}{2}=\frac{\text{y}-\frac{5}{2}}{-\frac{3}{2}}=\frac{\text{z}-2}{0}$

$\frac{\text{x}-2}{4}=\frac{\text{y}-\frac{5}{2}}{-3}=\frac{\text{z}-2}{0}$

The direction ratios of the given line are proportional to 4, -3, 0.

Hence, the direction cosines of the given line are proportional to

$\frac{4}{\sqrt{4^2+(-3)^2+0^2}},\frac{-3}{\sqrt{4^2+(-3)^2+0^2}},\frac{0}{\sqrt{4^2+(-3)^2+0^2}}$

$=\frac{4}{5},\frac{-3}{5},0$