Question 15 Marks
If O be the origin and the coordinates of P be (1, 2, -3), then find the equation of the plane passing through P and perpendicular to OP.
AnswerThe normal is passing through the points O(0, 0, 0) and P(1, 2, -3) So,
$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
Since the plane passes through P(1, 2, -3), $\vec{\text{a}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$ in the relation, we get
$\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=1+4+9$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=14$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=14$
Substituting $\overrightarrow{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=14$
$\Rightarrow\text{x}+2\text{y}-3\text{z}=14$
View full question & answer→Question 25 Marks
Find the equation of the plane through the points $(2, 2, -1)$ and $(3, 4, 2)$ and parallel to the line whose direction ratios are $7, 0, 6.$
AnswerThe equation of the plane through $(2, 2, -1)$ is
$a(x - 2) + b(y - 2) + c(z + 1) = 0 ....(i)$
This plane passes through $(3, 4, 2).$
so, $a(3 - 2) + b(4 - 2) + c(2 + 1) = 0$
$\Rightarrow a + 2b + 3c = 0 ....(ii)$
Again plane $(i)$ is parallel to the line whose direction ratios are $7, 0, 6.$
It means that the normal of plane $(i)$ is perpendicular to the line whose direction ratios are $7, 0, 6$
$\Rightarrow 7a + 0b + 6c = 0 ($Because $a_1a_2 + b_1b_2 + c_1c_{2 }= 0)$
Solving $(i), (ii)$ and $(iii),$ we get
$\begin{vmatrix}\text{x}-2&\text{y}-2&\text{z}+1\\1&2&3\\7&0&6\end{vmatrix}=0$
$\Rightarrow 12(x - 2) + 15(y - 2) - 14(z + 1) = 0$
$\Rightarrow 12x + 15y - 14z - 68 = 0$
View full question & answer→Question 35 Marks
Find the angle between the lines whose direction ratios are proportional to a, b, c and b - c, c - a, a - b.
AnswerGiven, that the direction ratios of lines are proportional to a, b, c and b - c, c - a, a - b.
Let, $\vec{\text{x}}$ and $\vec{\text{y}}$ be the vector parallel to these lines respectively, so
$\vec{\text{x}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$
And, $\vec{\text{y}}=(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}$
Let, $\theta$ be the angle between $\vec{\text{x}}$ and $\vec{\text{y}}$, so,
$\cos\theta=\frac{\vec{\text{x}}\times\vec{\text{y}}}{\big|\vec{\text{x}}\big|\big|\vec{\text{y}}\big|}$
$=\frac{(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}})[(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}]}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}}$
$=\frac{(\text{a})(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})+\text{c}(\text{a}-\text{b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2\sqrt{\text{b}^2+\text{c}^2-2\text{bc}+\text{c}^2+\text{a}^2-2\text{ac}+\text{a}^2+\text{b}^2-2\text{ab}}}}$
$\cos\theta=\frac{\text{ab}-\text{ac}+\text{bc}-\text{ba}+\text{ca}-\text{bc}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{2\text{a}^2+2\text{b}^2+2\text{c}^2-2\text{ab}-2\text{bc}-2\text{ca}}}$
$\cos\theta=0$
$\theta=\cos^{-1}(0)$
$\theta=\frac{\pi}{2}$
Angle between the lines $=\frac{\pi}{2}$
View full question & answer→Question 45 Marks
Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes $\text{x + 2y +3z = 5 and 3x + 3y + z} = 0$
Answer$\text{Let} \overrightarrow{a} = -\hat{\text{i}} + 3\hat{\text{j}} + 2\hat{\text{k}}$
Plane is perpendicular to the planes $\text{x + 2y + 3z = 5 and 3x + 3y + z = 0} $
$\therefore \text{normal to the plane is} \bigg( \hat{\text{i}} + 2\hat{\text{j}} + 3\hat{\text{k}}\bigg) \times \bigg(3{\text{i}} + 3\hat{\text j} + \hat{\text{k}}\bigg )$
$ = -7\hat{\text{i}} + 8\hat{\text{j}} - 3\hat{\text{k}}\text{ or } 7\hat{\text{i}} - 8\hat{\text{j}} + 3\hat{\text{k}}$
$\therefore \text{Equation of plane is}$
$\overrightarrow{r}. \bigg(7\hat{\text{i}} - 8\hat{\text{j}} + 3\hat{\text{k}}\bigg) = \bigg( \hat{\text{-i}} + 3\hat{\text{j}} + 2\hat{\text{k}}\bigg). \bigg(7\hat{\text{i}} - 8\hat{\text{j}} + 3\hat{\text{k}}\bigg) =- 25$
$\text{or 7x - 8y + 3z + 25 = 0}$
View full question & answer→Question 55 Marks
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x - y + z + 1 = 0. Also, find the image of the point in the plane.
Answer2x - y + z + 1 = 0
(3, 2, 1)
$=\Big|\frac{6-2+1+1}{\sqrt{4+1+1}}\Big|=\frac{6}{\sqrt{6}}=\sqrt{6}$
Let the foot of perpendicular be (x, y, z). So, DR's are in proportional
$\frac{\text{x}-3}{2}=\frac{\text{y}-2}{-1}=\frac{\text{z}-1}{1}=\text{k}$
$\text{x}=2\text{k}+3$
$\text{y}=-\text{k}+2$
$\text{z}=\text{k}-1$
Subsititute (x, y, z) = (2k + 3, -k + 2, k - 1) in plane equation
2x - y + z + 1 = 0
4k + 6 + k - 2 + k - 1 + 1 = 0
6k = -4
$\text{k}=\frac{-4}{6}=\frac{-2}{3}$
$(\text{x},\text{y},\text{z})=\Big(\frac{5}{3},\frac{8}{3},\frac{-5}{3}\Big)$
View full question & answer→Question 65 Marks
Show that the points $(2, 3, 4), (-1, -2, 1), (5, 8, 7)$ are collinear.
AnswerSuppose the points are $A(2, 3, 4), B(-1, -2, 1)$ and $C(5, 8, 7).$
We know that the direction ratios of the line joining the points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are $x_{2 }- x_{1, }y_{2 }- y_{1, }z_{2 }- z_{1.}$
The direction ratios of $AB$ are $(-1 - 2), (-2 - 3), (1 - 4),$
i.e. $-3, -5, -3.$
The direction ratios of $BC$ are $(5 - (-1)), (8 - (-2)), (7 - 1),$
i.e. $6, 10, 6.$
It can be seen that the direction ratios of $BC$ are $-2$ times that of $AB,$ i.e. they are proportional.
Therefore, $AB$ is parallel to $BC.$
Since point $B$ is common in both $AB$ and $BC,$ points $A, B,$ and $C$ are collinear.
View full question & answer→Question 75 Marks
Find the vector equations of the coordinate planes.
AnswerWe have to find vector equation of coordinate planes.
For xy-plane.
It passes through origin and is perpendicular to z-axis, so
Put $\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{k}}$ in the vector equation of plane passing through point $\vec{\text{a}}$ and perpendicular to vector $\vec{\text{n}}$
$(\vec{\text{r}}-\vec{\text{n}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{k}}=0$
$\vec{\text{r}}\cdot\vec{\text{k}}=0\ ...(\text{i})$
For xz-plane,
It passes throught origin and perpendicular to y-axis, so
$\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$ and $\vec{\text{n}}=\hat{\text{j}}$
Equation of xz-plane is given by
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{j}}=0$
$\vec{\text{r}}\cdot\hat{\text{j}}=0$
For yz-plane,
It passes throught origin and is perpendicular to x-axis, so
$\vec{\text{a}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}},\vec{ \text{n}}=\hat{\text{i}}$
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0$
$(\vec{\text{r}}-0\hat{\text{i}}-0\hat{\text{j}}-0\hat{\text{k}})\hat{\text{i}}=0$
$\vec{\text{r}}\cdot\hat{\text{i}}=0$
Hence, equation of xy, yz, zx-plane are given by
$\vec{\text{r}}\cdot\vec{\text{k}}=0$
$\vec{\text{r}}\cdot\hat{\text{i}}=0$
$\vec{\text{r}}\cdot\hat{\text{j}}=0$
View full question & answer→Question 85 Marks
Find the vector equation of the following planes in non-parametric form.
$\vec{\text{r}}=(2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})+\lambda(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\mu(5\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}})$
AnswerWe know that equation $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}+\mu\vec{\text{c}} $ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to t.Here, $\vec{\text{a}}=2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},\hat{\text{c}}=5\hat{\text{i}}-2\hat{\text{j}}+7\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&2&3\\5&-2&7\end{vmatrix}$
$=20\hat{\text{i}}+8\hat{\text{j}}-12\hat{\text{k}}$
The vector equation of the plane in scalar product form is,
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(20\hat{\text{i}}+8\hat{\text{j}}-12\hat{\text{k}})=(2\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})(20\hat{\text{i}}+8\hat{\text{j}}-12\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\big(4(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\big)=40+16+12$
$\Rightarrow\vec{\text{r}}\cdot\big(4(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})\big)=68$
$\Rightarrow\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}})=17$
View full question & answer→Question 95 Marks
Find the cartesian form of the equations of the following planes.
$\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}})+\text{s}(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})+\text{t}(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$
Answer $\vec{\text{r}}=(\hat{\text{i}}-\hat{\text{j}})+\text{s}(-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})+\text{t}(\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}})$We know that the equation $\vec{\text{r}}=\vec{\text{a}}+\text{s}\vec{\text{b}}+\text{t}\vec{\text{c}}$ represents a plane passing through a point whose position vector is $\vec{\text{a}}$ and parallel to the vectors $\vec{\text{b}}$ and $\vec{\text{c}}.$
Here, $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}},\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&1&2\\1&2&1\end{vmatrix}$
$=-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}}$
The vector equation of the plane in scalar product form is,
$\vec{\text{i}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})=(\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}})\cdot(-3\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[-3(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\Big]=-3-3+0 $
$\Rightarrow\vec{\text{r}}\cdot\Big[-3(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})\Big]=-6$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=2$
For cartesian form, let us substitute $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ here. Then we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}})=2$
$\text{x}-\text{y}+\text{z}=2$
View full question & answer→Question 105 Marks
Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}$ and $\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}$
Answer$\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}\dots(1)$$\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}\dots(2)$
Since line (1) passes through the point ( 3, 5, 7) and has direction ratios proportional to 1, -2, 1, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$
Here,
$\vec{\text{a}}_1=3\hat{\text{i}}+5\hat{\text{j}}+7\hat{\text{k}}$
$\vec{\text{b}}_1=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Also, line (2) passes through the point (-1, -1, -1) and has diraction ratios proportional to 7, -6, 1.
Its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$
Here,
$\vec{\text{a}}_2=-\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$
$\vec{\text{b}}_2=7\hat{\text{i}}-6\hat{\text{j}}+\hat{\text{k}}$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-4\hat{\text{i}}-6\hat{\text{j}}-8\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-2&1\\7&-6&1\end{vmatrix}$
$=4\hat{\text{i}}+6\hat{\text{j}}+8\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{4^2+6^2+8^2}$
$=\sqrt{16+36+64}$
$=\sqrt{116}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-4\hat{\text{i}}-6\hat{\text{j}}-8\hat{\text{k}}\big).\big(4\hat{\text{i}}+6\hat{\text{j}}+8\hat{\text{k}}\big)$
$=-16-36-64$
$=-116$
The shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{-116}{\sqrt{116}}\Big|$
$=\sqrt{116}$
$=2\sqrt{29}$
View full question & answer→Question 115 Marks
Find the equation of the plane which contains planes is the line of intersection of the planes x + 2y + 3z - 4 = 0 and 2x + y - z + 5 = 0 and whose x-intercept is twice its z-intercept.
AnswerHence write the vector equation of a plane passing through the point (2, 3, -1) and parallel to the plane obtained above.Solution:
Given, eqcation of planes are
$\text{x}+2\text{y}+3\text{z}-4=0$ . . .(i)
and $2\text{x}+\text{y}+\text{z}+5=0$ . . . (ii)
Clearly, the equation of the plan which contains the line of intersection of planes (i)and (ii), is
$(\text{x}+2\text{y}+3\text{z}-4)+\lambda(2\text{x}+\text{y}+\text{z}+5)=0$
or $(1+2\lambda)\text{x}+(2+\lambda)\text{y}+(3-\lambda)\text{z}+5\lambda-4=0$ . . .(iii)
This equation can be written in intercept form as
$\frac{\text{x}}{\frac{4-5\lambda}{1+2\lambda}}+\frac{\text{y}}{\frac{4-5\lambda}{2+\lambda}}+\frac{\text{z}}{\frac{4-5\lambda}{3+\lambda}}=1$
Since, it is given that the x-intercept of plane (iii) is twice its z-intercept.
$\therefore \frac{4-5\lambda}{1+2\lambda}=2\bigg(\frac{4-5\lambda}{3-\lambda}\bigg) $
or $3-5\lambda=2+4\lambda$
or $5\lambda=1 $or $ \lambda=\frac{1}{5}$
So,the required equation of plane is
$\bigg(1+\frac{2}{5}\bigg)\text{x}+\bigg(2+\frac{1}{5}\bigg)\text{y}+\bigg(3-\frac{1}{5}\bigg)\text{z}=4-5.\frac{1}{5} $
or $\frac{7}{5}\text{x}+\frac{11}{5}\text{y}+\frac{14}{5}\text{z}=\frac{15}{5}$
or $7\text{x}+11\text{y}+14\text{z}=15$ ...(iii)
Clearly the Dr's of normal to the plane , which is parallel to the plane (iv), are 7, 11 and 14.
$\therefore$ The vector equation of a plane passing through the (2, 3, -1) and parallel to the plane (iv), is
$\big[\overrightarrow{\text{r}}-(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\big].(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=0$
or $\overrightarrow{\text{r}}.(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}).(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})$
or $\overrightarrow{\text{r}}(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=14+33-14$
or $\overrightarrow{\text{r}}(7\hat{\text{i}}+11\hat{\text{j}}+\hat{\text{14k}})=33$\
View full question & answer→Question 125 Marks
Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)+\mu\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)$
Answer$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)+\mu\big(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}\big)$Comparing the given equations with the equations $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ we get
$\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$
$\vec{\text{a}}_2=2\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$
$\vec{\text{b}}_1=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}$
$\therefore\vec{\text{a}}_2-\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{J}}&\hat{\text{k}}\\2&3&4\\3&4&5 \end{vmatrix}$
$=-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\Big|=\sqrt{(-1)^2+2^2+(-1)^2}$
$=\sqrt{1+4+1}$
$=\sqrt{6}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}\big).\big(-\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)$
$=-1+4-2$
$=1$
The shrtest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$ is given by
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{1}{\sqrt{6}}\Big|$
$=\frac{1}{\sqrt{6}}$
View full question & answer→Question 135 Marks
Determine whether the following pair of lines intersect or not:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\text{z}$ and $\frac{\text{x}+1}{5}=\frac{\text{y}-2}{1};\text{z}=2$
AnswerGiven, equation of first line is$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\frac{\text{}z}{1}=\lambda\text{ (say) }\dots(1)$
General point on line (1) is
$\big(2\lambda+1,3\lambda-1,\lambda\big)$
Another equation of line is
$\frac{\text{x}-1}{5}=\frac{\text{y}-2}{1},\text{z}=3\dots(2)$
$\frac{\text{x}-5}{5}=\frac{\text{y}-2}{1}=\mu,\text{ (say) },\text{z}=3$
General point on line (2) is
$\big(5\mu+1,\mu+2,3\big)$
If line (1) and (2) intersect each other then, there is a common point to them, so, we must have of $\lambda$ and $\mu$ such that
$2\lambda+1=5\mu+1\Rightarrow2\lambda-5\mu=0\dots(3)$
$3\lambda-1=\mu+2\Rightarrow3\lambda-\mu=3\dots(4)$
$\lambda=3\Rightarrow\lambda=3\dots(5)$
Put value of $\lambda$ in equation (4),
$3\lambda-\mu=3$
$3(3)-\mu=3$
$-\mu=3-9$
$\mu=6$
Put the value of $\lambda$ and $\mu$ in equation (3), so
$2\lambda-5\mu=0$
$2(3)-5(6)=0$
$6-30=0$
$-24\neq0$
$\text{LHS}\neq\text{RHS}$
sice the values of $\lambda$ and $\mu$ obtained from equation (4) and (5) dose not satisfy equation (3),
So, given lines are not intersecting.
View full question & answer→Question 145 Marks
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P(3, 2, 1) from the plane 2x – y + z + 1 = 0. Find also, the image of the point in the plane.
Answer
Let Q be the foot of perpendicular from P to the plane
$\therefore\text{Equation of PQ is }\frac{\text{x - 3}}{2}=\frac{\text{y - 2}}{-1}=\frac{\text{z - 1}}{1}$
Any point on this line is $(2\lambda+3,-\lambda+2,\lambda+1)$
If this point is Q, then it must satisfy the equation of plane
$\therefore2(2\lambda+3)-(-\lambda+2)+(\lambda+1)+1=0$
$\Rightarrow\lambda=-1$
$\therefore$ coordinates of foot of perpendiculare are Q (1, 3, 0)
Perpendicular distance = PQ = $\sqrt{4+1+1}=\sqrt{6}\text{ units}$
Let P' (x, y, z) be the image, then $\frac{\text{x + 3}}{2}=1,\frac{\text{y + 2}}{2}=3, \frac{\text{z + 1}}{2}=0$
$\therefore \text{P' is }(-1, 4, -1)$. View full question & answer→Question 155 Marks
If the axes are rectangular and p is the point (2, 3, -1), find the equation of the plane throught p at right angles to OP.
AnswerThe normal is passing through the points O(0, 0, 0) and P(2, 3, -1). So,$\vec{\text{n}}=\overrightarrow{\text{OP}}$
$=(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})-(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}})$
$=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
Since the plane passes through the point (2, 3, -1)
$=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$
We know that the vector equation of the plane passing through a point $\vec{\text{a}}$ and normal to $\vec{\text{n}}$ is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
Substituting $\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{n}}=2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}},$ we get
$\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=4+9+1$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
Substituting $\vec{\text{r}}=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$ in the vector equation, we get
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=14$
$\Rightarrow2\text{x}+3\text{y}-\text{z}=14$
View full question & answer→Question 165 Marks
Prove that the line through A(0, –1, –1) and B(4, 5, 1) intersects the line through C(3, 9, 4) and D(–4, 4, 4).
AnswerEquation of line $\vec{\text{AB}}$
$\vec{\text{r}} = (-\hat{\text{j}} - \hat{\text{k}}) + \lambda (4\hat{\text{i}} + 6\hat{\text{j}} + 2\hat{\text{k}})$
Equation of line $\vec{\text{CD}}$
$\vec{\text{r}} = (3\hat{\text{i}} + 9\hat{\text{j}} + 4\hat{\text{k}}) + \mu (-7\hat{\text{i}} - 5\hat{\text{j}})$
$\vec{\text{a}}_{2} - \vec{\text{a}}_{1} = 3\hat{\text{i}} + 10\hat{\text{j}} + 5 \hat{\text{k}}$
$\vec{\text{b}}_{1} \times \vec{\text{b}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & 6 & 2 \\ -7 & -5 & 0 \end{vmatrix} = 10\hat{\text{i}} - 14\hat{\text{j}} + 22\hat{\text{k}}$
$(\vec{\text{a}}_{2} - \vec{\text{a}}_{1}). (\vec{\text{b}}_{1} \times \vec{\text{b}}_{2}) = 30 – 140 + 110 = 0$
$\Rightarrow$ Lines intersect.
View full question & answer→Question 175 Marks
Find the perpendicular distence of the point (1, 0, 0) from the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}.$ Also, find the coordinates of the perpendicular and the equation of the perpendicular.
AnswerLet foot of the perpemdicular drawn from the point P(1, 0, 0) to the line $\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}$ is Q. we have to find lengh of PQ.
Q is a genelar point on the line,
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{-3}=\frac{\text{z}+10}{8}=\lambda$ (say)
coordinate of Q $=\big(2\lambda+1,-3\lambda-1,8\lambda-10\big)$
Direction ratios line PQ are
$=\big(2\lambda+1-1\big),\big(-3\lambda-1-0\big),\big(8\lambda-10-0\big)$
$=(2\lambda),\big(-3\lambda-1\big),\big(8\lambda-10\big)$
Since, line PQ is perpendicular to the given line, so
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$(2)(2\lambda)+(-3)(-3\lambda-1)+8(8\lambda-10)=0$
$4\lambda+9\lambda+3+64\lambda-80=0$
$77\lambda-77=0$
$\lambda=1$
Therefore, coordinate of Q is $\big(2\lambda+1,-3\lambda-1,8\lambda-10\big)$
$=\big(2(1)+1,-3(1)-1,8(1)-10\big)$
$=(3,-4,-2)$
$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{(1-3)^2+(0+4)^2+(0+2)^2}$
$=\sqrt{4+16+4}$
$=\sqrt{24}$
$=2\sqrt{6}$
So, foot of perpendicular $=(3,-1,-2)$
length of perpendicular $=2\sqrt{6}\text{ units}$
View full question & answer→Question 185 Marks
Show that the plane whose vector equation is $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the line whose vector equation is $\vec{\text{r}}=\hat{\text{i}}+\hat{\text{j}}+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}).$
AnswerThe line $\vec{\text{r}}=(\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}})+\lambda(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\ ...(\text{i})$
Passes through a point whose posotion vector is $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}}$
If the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})=3$ contains the given line, then
It should passes through the point $\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}.$
It should be parallel to the line.
Now, the plane passes through the point $\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}.$
So, the plane vector to the given plane is $\vec{\text{n}}=\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}.$
We observe that
$\vec{\text{b}}\cdot\vec{\text{n}}=(2\hat{\text{i}}+\hat{\text{j}}+4\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}})$
$=2+2-4=0$
Therefore, the plane is parallel to the line.
Hence, the given plane contains the given line.
View full question & answer→Question 195 Marks
Find the equation of the plane through the points $(2, 1, 0), (3, -2, -2)$ and $(3, 1, 7).$
AnswerWe know that, the equation of a plane passing through three non-collinear points $(x_1, x_1, x_1), (x_2, x_2, x_2)$ and $(x_3, x_3, x_3)$ is
$\begin{vmatrix}\text{x}-\text{x}_1&\text{y}-\text{y}_1&\text{z}-\text{z}_1 \\\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{x}_3-\text{x}_1&\text{y}_3-\text{y}_1&\text{z}_3-\text{z}_1 \end{vmatrix}=0$
$\Rightarrow\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} \\1&-3&-2\\1&0&7 \end{vmatrix}=0$
$\Rightarrow(\text{x}-2)(-21+0)-(\text{y}-1)(7+2)+\text{z}(3)=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$
$\therefore7\text{x}+3\text{y}-\text{z}=17$
So, the required equation of plane is $7\text{x}+3\text{y}-\text{z}=17.$
View full question & answer→Question 205 Marks
Find the coordinates of the point where the line through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $ZX-$plane.
AnswerGiven: A line through the points $A(5, 1, 6)$ and $B(3, 4, 1)$
$\therefore$ Direction ratios of this line $AB$ are $x_2 - x_1, y_2 - y_1, z_2 - z_1$
$\Rightarrow 3 - 5, 4 - 1, 1 - 6$
$\Rightarrow -2, 3, -5 = a, b, c$
Equation of the line $AB$ is $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{\text{y}-1}{3}=\frac{\text{z}-6}{-5}\ \ \ .....(\text{i})$
Now we have to find the coordinates of the point where this line $AB$ crosses the $ZX-$plane
i.e., $y = 0 .......(ii)$
Putting $y = 0$ in eq. $(i),$ we get
$\frac{\text{x}-5}{-2}=\frac{-1}{3}=\frac{\text{z}-6}{-5}$
$\Rightarrow\ \ \frac{\text{x}-5}{-2}=\frac{-1}{3}$ and $\frac{\text{z}-6}{-5}=\frac{-1}{3}$
$\Rightarrow 3x - 15 = 2$ and $3z - 18 = 5$
$\Rightarrow 3x = 17$ and $3z = 23$
$\Rightarrow\ \ \ \text{x}=\frac{17}{3}$ and $\text{z}=\frac{23}{3}$
Thus, required point is $\text{P}\Big(\ \frac{17}{3},\ 0,\frac{23}{3}\Big).$
View full question & answer→Question 215 Marks
Find the angle between the follwing pairs of lines:$\vec{\text{r}}=\lambda\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{j}}+\mu\big\{\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}$
Answer$\vec{\text{r}}=\lambda\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$ and $\vec{\text{r}}=2\hat{\text{j}}+\mu\big\{\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}\big\}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}$
If $\theta$ is the angle between the given 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{\big(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big).\big(\big(\sqrt{3}-1\big)\hat{\text{i}}-\big(\sqrt{3}+1\big)\hat{\text{j}}+4\hat{\text{k}}\big)}{\sqrt{1^2+1^2+2^2}\sqrt{(\sqrt{3}-1)^2+(\sqrt{3}+1)^2+4^2}}$
$=\frac{\big(\sqrt{3}-1\big)-\big(\sqrt{3}+1\big)+8}{\sqrt{6}\sqrt{24}}$
$=\frac{6}{12}$
$=\frac{1}{2}$
$\Rightarrow\theta=\frac{\pi}{3}$
View full question & answer→Question 225 Marks
A(1, 0, 4) B(0, -11, 1), C(2, -3, 1) are three points and D is the fool of perpendicular from A on BC. Find the coordinates of D.
AnswerPoint D is the foot of the perpendicular drawn from the point A(1, 0, 4) to the line BC.
The coordinates of a general point on the line BC are given by
$\frac{\text{x}-0}{2-0}=\frac{\text{y}+11}{-3+11}=\frac{\text{z}-3}{1-3}=\lambda$
$\Rightarrow\text{x}=2\lambda$
$\text{y}=8\lambda-11$
$\text{z}=-2\lambda+3$
Let the coordinates of D be $(2\lambda,8\lambda-11,-2\lambda+3).$
The direction ratios of AD are proportional to
$2\lambda-1,8\lambda-11,-2\lambda+3-4,$
i.e. $2\lambda-1,8\lambda-11,-2\lambda-1.$
The direction ratios of the line BC are proportional to 2, 8 -2, but AD is perpendicular to the line BC.
$\therefore2(2\lambda-1)+8(8\lambda-11)-2(-2\lambda-1)=0$
$\Rightarrow\lambda=\frac{11}9{}$
Substituting $\lambda=\frac{11}{9}$ in $(2\lambda,8\lambda-11,-2\lambda+3),$ we get the coordinates of D as $\Big(\frac{22}{9},-\frac{11}{9},\frac{5}{9}\Big).$
View full question & answer→Question 235 Marks
Find the equation of the plane which is perpendicular to the plane $5x + 3y + 6z + 8 = 0$ and which contains the line of intersection of the planes $x + 2y + 3z - 4 = 0$ and $2x + y - z + 5 =0$
AnswerEquation of plane, containing the line of intersection of Planes:$ x + 2y + 3z - 4 = { 0 } \text{ and 2x} + y - z + 5 = \text{ 0 is}$
${(x + 2y + 3z - 4)} + \lambda {(2x + y - z + 5}) = 0$
$\Rightarrow {x (1 + 2\lambda ) + y(2 + \lambda) + z (3 - \lambda) + ( - 4 + 5\lambda )} = 0 \dots\dots\dots\dots\text{(i})$
$\text{(i) is} \perp \text{to 5x} + 3y + 6z + 8 = 0$
$\therefore 5 ( 1 + 2\lambda ) + 3 ( 2 + \lambda ) + 6 ( 3 - \lambda) = 0 \Rightarrow \lambda = - \frac{29}{7}$
$\therefore$ Requuired equation of plane is:
$x \bigg( 1 - \frac{58}{7}\bigg) + y \bigg(2 -\frac{29}{7}\bigg)+\text{z}\Big(3+\frac{29}{7}\Big)- 4 - \frac{145}{7} = 0$
$\Rightarrow 51 x + 15 y- 50 z + 173 = 0$
View full question & answer→Question 245 Marks
Find the coordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, -1, 3) and C(2, -3, -1).
AnswerThe cartesian equation of the line joining points B(0, -1, 3) and C(2, -3, -1) is
$\frac{\text{x}-0}{2-0}=\frac{\text{y}-(-1)}{-3-(-1)}=\frac{\text{z}-3}{-1-3}$
Or $\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}$
Let L be the foot of the perpendicular drawn from the point A(1, 8, 4) to the line $\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}.$
The coordinates of general point on the line $\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}$ are given by
$\frac{\text{x}}{2}=\frac{\text{y}+1}{-2}=\frac{\text{z}-3}{-4}=\lambda$
Or $\text{x}=2\lambda,\text{y}=-2\lambda-1,\text{z}=-4\lambda+3$
Let the coordinates of L be $(2\lambda,-2\lambda-1,-4\lambda+3).$ Therefore, the direction ratios of AL are proportional to
$2\lambda-1,-2\lambda-1-8,-4\lambda+3-4$ or $2\lambda-1,-2\lambda-9,-4\lambda-1$
Direction ratios of the given line are proportional to 2, -2, -4.
But, AL is perpendicular to the given line.
$\therefore2\times(2\lambda-1)+(-2)\times(-2\lambda-9)+(-4)\times(-4\lambda-1)=0$
$\Rightarrow4\lambda-2+4\lambda+18+16\lambda+4=0$
$\Rightarrow24\lambda+20=0$
$\Rightarrow\lambda=-\frac{5}{6}$
Puting $\lambda=-\frac{5}{6} $ in $(2\lambda,-2\lambda-1,-4\lambda+3),$ we get
$\Big(2\times\Big(-\frac{5}{6}\Big),-2\times\Big(-\frac{5}{6}\Big)-1,-4\times\Big(-\frac{5}{6}\Big)+3\Big)=\Big(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\Big)$
Thus, the required coordinates of the foot of the perpendicular are $\Big(-\frac{5}{3},\frac{2}{3},\frac{19}{3}\Big).$
View full question & answer→Question 255 Marks
Find the angle between the pairs of lines with direction ratios proportional to $1, 2, -2$ and $-2, 2, 1$
AnswerWe know that, angle $(\theta)$ between two lines$\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ and $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$
is given by,
$\cos\theta=\frac{\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2}{\sqrt{{\text{a}_1}^2+{{\text{b}_1}^2}+{{\text{c}_1}^2}}\sqrt{{\text{a}_2}^2+{{\text{b}_2}^2}+{{\text{c}_2}^2}}}\dots(1)$
Here, $a_{1 }= 1, b_{1 }= 2, c_{1 }= -2$
$a_{2 }= -2, b_{2 }= 2, c_{2 }= 1$
Let $\theta$ be required angle, so using equation $(1),$
$\cos\theta=\frac{(1)(-2)+(2)(2)+(-2)(1)}{\sqrt{(1)^2+(2)^2+(-2)^2}\sqrt{(-2)^2+(2)^2+(1)^2}}$
$=\frac{-2+4-2}{3.3}$
$=\frac{0}{9}$
$\cos\theta=0$
$\theta=\frac{\pi}{2}$
View full question & answer→Question 265 Marks
Find the angle between the following pairs of lines:$\frac{-\text{x}+2}{-2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{2\text{y}-8}{4}=\frac{\text{z}-5}{4}$
Answer$\frac{-\text{x}+2}{-2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{2\text{y}-8}{4}=\frac{\text{z}-5}{4}$
The equation of the given lines can be re-written as
$\frac{\text{x}-2}{2}=\frac{\text{y}-1}{7}=\frac{\text{z}+3}{-3}$ and $\frac{\text{x}+2}{-1}=\frac{\text{y}-4}{2}=\frac{\text{z}-5}{4}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given line.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}_2=-1\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}$
If $\theta$ is the angle between the given 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{\big(2\hat{\text{i}}+7\hat{\text{j}}-3\hat{\text{k}}\big).\big(-1\hat{\text{i}}+2\hat{\text{j}}+4\hat{\text{k}}\big)}{\sqrt{2^2+7^2+(-3)^2}\sqrt{(-1)^2+2^2+4^2}}$
$=\frac{-2+14-12}{\sqrt{62}\sqrt{21}}$
$=0$
$\Rightarrow\theta=\frac{\pi}{2}$
View full question & answer→Question 275 Marks
$\text{If }\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{7k}};\text{ and }\overrightarrow{b}=\hat{\text{5i}}-\hat{\text{j}}+\lambda\hat{\text{k}},$ then find the value of $\lambda,$ so that $\overrightarrow{a}+\overrightarrow{b}$ and $\overrightarrow{a}-\overrightarrow{b}$ are perpendicular vectors.
AnswerHere $\text{If }\overrightarrow{\text{a}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{7k}};\text{ }\overrightarrow{b}=\hat{\text{5i}}-\hat{\text{j}}+\lambda\hat{\text{k}},$$\therefore \overrightarrow{a}+\overrightarrow{b}=6\hat{\text{i}}-2\hat{\text{j}}+\text{(7}+\lambda)\hat{\text{k}};\text{ } \overrightarrow{a}-\overrightarrow{b}=-4\hat{\text{i}}+\text{(7}-\lambda)\hat{\text{k}}$
$\therefore\text{ }\overrightarrow{(a}+\overrightarrow{b)}\text{ is perpendicular to }\overrightarrow{(a}-\overrightarrow{b)}$
$\Rightarrow \text{ }\overrightarrow{(a}+\overrightarrow{b)}.\overrightarrow{(a}-\overrightarrow{b)}=0\text{ }\Rightarrow \text{ -24 + (7}+\lambda).\text{(7} - \lambda)=0$
$\Rightarrow$ –24 + 49 – $\lambda^2 $ = 0 $\Rightarrow$ $\lambda^2 $ = 25
$\Rightarrow$ $\lambda $ = ± 5.
View full question & answer→Question 285 Marks
Find the angle between the pairs of lines with direction ratios proportional toa, b, c and b - c, c - a, a - b.
Answera, b, c and b - c, c - a, a - b are direction ratios these are the vectors with above direction ratios $\hat{\text{x}}=\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}},\hat{\text{y}}=(\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}$ are the vectors parallel to two given lines $\therefore$ angle between the lines with above direction ratios are $\hat{\text{x}}$ and $\hat{\text{y}}\rightarrow\cos\theta=\frac{\hat{\text{x}}.\hat{\text{y}}}{|\hat{\text{x}}||\hat{\text{y}}|}$ $\cos\theta=\frac{\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big).\big((\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}\big)}{\big|\big(\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}\big)\big|\big|\big((\text{b}-\text{c})\hat{\text{i}}+(\text{c}-\text{a})\hat{\text{j}}+(\text{a}-\text{b})\hat{\text{k}}\big)\big|}$ $=\frac{\text{a}(\text{b}-\text{c})+\text{b}(\text{c}-\text{a})+\text{c}(\text{a}-\text{b})}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}$ $=\frac{\text{ab}-\text{ac}+\text{bc}-\text{ba}+\text{ca}-\text{cb}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}\sqrt{(\text{b}-\text{c})^2+(\text{c}-\text{a})^2+(\text{a}-\text{b})^2}}=0$$\cos\theta=0\rightarrow\theta=90^\circ$
View full question & answer→Question 295 Marks
Find the angle between the following pairs of lines:$\frac{5-\text{x}}{-2}=\frac{\text{y}+3}{1}=\frac{1-\text{z}}{3}$ and $\frac{\text{x}}{3}=\frac{1-\text{y}}{-2}=\frac{\text{z}+5}{-1}$
Answer$\frac{5-\text{x}}{-2}=\frac{\text{y}+3}{1}=\frac{1-\text{z}}{3}$ and $\frac{\text{x}}{3}=\frac{1-\text{y}}{-2}=\frac{\text{z}+5}{-1}$The equations of the given lines can be re-written as
$\frac{\text{x}-5}{2}=\frac{\text{y}+3}{1}=\frac{\text{z}-1}{-3}$ and $\frac{\text{x}}{3}=\frac{\text{y}-1}{2}=\frac{\text{z}+5}{-1}$
Let $\vec{\text{b}}_1$ and $\vec{\text{b}}_2$ be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}$
If $\theta$ is the angle between the given 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{\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big).\big(3\hat{\text{i}}+2\hat{\text{j}}-\hat{\text{k}}\big)}{\sqrt{2^2+1^2+(-3)^2}\sqrt{3^2+2^2+(-1)^2}}$
$=\frac{6+2+3}{\sqrt{14}\sqrt{14}}$
$=\frac{11}{14}$
$\Rightarrow\theta=\cos^{-1}\Big(\frac{11}{14}\Big)$
View full question & answer→Question 305 Marks
Find the equation of the line passing through the points $\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}$ and perpendicular to the lines $\vec{\text{r}}=\hat{\text{i}}+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).$
AnswerWe know that equation of a line passing through a point with position vector $\vec{\text{a}}$ and perpendiculat to $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)\dots(1)$
Here, $\vec{\text{a}}=\big(\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)$
and required line is perpendicular to
$\vec{\text{r}}=\hat{\text{i}}+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)$ and
$\vec{\text{r}}=\big(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\Rightarrow\vec{\text{b}}_1=\big(2\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big),\vec{\text{b}}_2=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Now,
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&1&-3\\1&1&1\end{vmatrix}$
$=\hat{\text{i}}(1+3)-\hat{\text{j}}(2+3)+\hat{\text{k}}(2-1)$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=4\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}$
Using equation, required equation of line is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$
$\vec{\text{r}}=\big(\hat{\text{i}}+\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(4\hat{\text{i}}-5\hat{\text{j}}+\hat{\text{k}}\big)$
View full question & answer→Question 315 Marks
Find the Cartesian equation of the plane passing through the points A(0, 0, 0) and B(3, –1, 2) and parallel to the line $\frac{\text{x-4}}{1}=\frac{\text{y+3}}{-4}=\frac{\text{z+1}}{7}$ .
AnswerEquation of plane passing through (0,0,0) is$\therefore$ a(x – 0) + b (y – 0) + c (z – 0) = 0 $\Rightarrow$ ax + by + cz = 0 .............(i)
It passes through (3, –1, 2)
$\therefore$ 3a – b + 2 c = 0 ........................(ii)
line $\frac{\text{x-4}}{1}=\frac{\text{y+3}}{-4}=\frac{\text{z+1}}{7}$ is | | to the plane (i)
$\Rightarrow$ a – 4 b + 7c = 0 .............................................(iii)
From (ii) and (iii), a = 1, b = – 19 and c = – 11
Equation of plane is x – 19y – 11z = 0.
View full question & answer→Question 325 Marks
Show that the lines $\frac{\text{x}+1}{3}=\frac{\text{y}+3}{5}=\frac{\text{z}+5}{7}$ and $\frac{\text{x}-2}{1}=\frac{\text{y}-4}{3}=\frac{\text{z}-6}{5}$ intersect. Find their point of intersection.
AnswerGiven equation of first line is $\frac{\text{x}+1}{3}=\frac{\text{y}+3}{5}=\frac{\text{z}+5}{7}=\lambda\text{ (say) }\dots(1)$ General point on line (1) is $\big(3\lambda-1,5\lambda-3,7\lambda-5\big)$ Another equation of line is $\frac{\text{x}-2}{1}=\frac{\text{y}-4}{3}=\frac{\text{z}-6}{5}=\mu\text{ (say) }\dots(2)$ General point on line (2) is, $\big(\mu+2,3\mu+4,5\mu+6\big)$ If lines (1) and (2) are intersecting then, they have a common point. So for same value of $\lambda$ and $\mu,$ we must have, $3\lambda-1=\mu+2\Rightarrow3\lambda-\mu=3\dots(3)$ $5\lambda-3=3\mu+4\Rightarrow5\lambda-3\mu=7\dots(4)$ $7\lambda-5=5\mu+6\Rightarrow7\lambda-5\mu=11\dots(5)$ Solving equation (3) and (4) to get $\lambda$ and $\mu,$
$\mu=\frac{3}{2}$ Put the value of $\mu$ in equation (3), $3\lambda-\mu=3$ $3\lambda-\Big(-\frac{3}{2}\Big)=3$ $3\lambda=3-\frac{3}{2}$ $\lambda=\frac{1}{2}$ put the value of $\lambda$ and $\mu$ in equation (5), $7\lambda-5\mu=11$ $7\Big(\frac{1}{2}\Big)-5\Big(-\frac{3}{2}\Big)=11$ $\frac{7}{2}+\frac{15}{2}=11$ $\frac{22}{2}=11$ $11=11$ $\text{LHS}\neq\text{RHS}$ Since, the values of $\lambda$ and $\mu$ obtained by solving (3) and (4) satisfy equation (5), Hence Given lines intersect each other. Point of intersection $=\big(3\lambda-1,5\lambda-3,7\lambda-5\big)$ $\Big\{\frac{3}{2}-1,\Big(\frac{5}{2}-3\Big),\Big(\frac{7}{2}-5\Big)\Big\}$ $=\Big(\frac{1}{2},\frac{-1}{2},\frac{-3}{2}\Big)$ Point of intersection is $\Big(\frac{1}{2},\frac{-1}{2},\frac{-3}{2}\Big).$ View full question & answer→Question 335 Marks
Show that the lines $\frac{\text{x}-1}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{5}$ and $\frac{\text{x}+2}{4}=\frac{\text{y}-1}{3}=\frac{\text{z}+1}{-2}$ do not intersect.
AnswerThe coordinates of any point on the first line are given by$\frac{\text{x}-1}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{5}=\lambda$
$\Rightarrow\text{x}=3\lambda+1$
$\text{y}=2\lambda-1$
$\text{z}=5\lambda+1$
The coordinates of a general point on the first line are $\big(3\lambda+1,2\lambda-1,5\lambda+1\big).$
The coordinates of any point on the second line are given by
$\frac{\text{x}+2}{4}=\frac{\text{y}-1}{3}=\frac{\text{z}+1}{-2}=\mu$
$\Rightarrow\text{x}=4\mu-2$
$\text{y}=3\mu+1$
$\text{z}=-2\mu-1$
The coordinates of a general point on the second line are $\big(4\mu-2,3\mu+1,-2\mu-1\big).$
If the lines intersect, then they have a common point. so, for some values of $\lambda$ and $\mu,$ we must have
$3\lambda+1=4\mu-2,2\lambda-1=3\mu+1,5\lambda+1=-2\mu-1$
$\Rightarrow3\lambda-4\mu=-3\dots(1)$
$2\lambda-3\mu=2\dots(2)$
$5\lambda+2\mu=-2\dots(3)$
Solving (1) and (2), we get
$\lambda=-17$
$\mu=-12$
Substituting $\lambda=-17$ and $\mu=-12$ in (3), we get
$\text{LHS}=3\lambda+2\mu$
= 3(-17) + 2( -12)
= -75
≠ -2
$\Rightarrow\text{LHS}\neq\text{RHS}$
Hence, the given lines do not intersect.
View full question & answer→Question 345 Marks
Find the shortest distance between the lines $\frac{\text{x}-1}{2}=\frac{\text{y}-3}{4}=\frac{\text{z}+2}{1}$ and 3x - y - 2z + 4 = 0 = 2x + y + z + 1.
AnswerThe equationg of the plane containing the line 3x - y - 2z + 4 = 0 = 2x + y + z + 1 is
$(3\text{x}-\text{y}-2\text{z}+4)+\lambda(2\text{x}+\text{y}+\text{z}+1)=0$
Or $(3+2\lambda)\text{x}+(\lambda-1)\text{y}+(\lambda-2)\text{z}+(\lambda+4)=0\ .....(\text{i})$
If it is parallel to the line $\frac{\text{x}-1}{2}=\frac{\text{y}-3}{4}=\frac{\text{x}+2}{1},$ then
$2(3+2\lambda)+4(\lambda-1)+(\lambda-2)=0$
$\Rightarrow 9\lambda = 0$
$\Rightarrow\ \lambda=0$
Putting $\lambda=0$ in (i), we get
3x - y - 2z + 4 = 0 ......(ii)
This is the equation of the plane containing the second line and parallel to the first line.
Now, the line $\frac{\text{x}-1}{2}=\frac{\text{y}-3}{4}=\frac{\text{x}+2}{4}$ passes through (1, 3, -2)
$\therefore$ Shortest distance between the given lines
= Lnegth of the perpendicular from (1, 3, -2) to the plane 3x - y - 2z + 4 = 0
$=\bigg|\frac{3\times1-3-2\times(-2)+4}{\sqrt{3^2+(-1)^2+(-2)^2}}\bigg|$
$=\bigg|\frac{3-3+4+4}{\sqrt{9+1+4}}\bigg|$
$=\frac{8}{\sqrt{14}}\text{ units}$
View full question & answer→Question 355 Marks
The vector equations of two lines are:$\overrightarrow{r} = \hat{i} +2\hat{j} +3\hat{k} +\lambda (\hat{i}-3\hat{j} +2\hat{k}) \text{and} \overrightarrow{r} = 4\hat{i} +5\hat{j} +6\hat{k} + \mu (2\hat{i}-3\hat{j} +\hat{k})$
Find the shortest distance between the above lines.
Answer$\text{Here} \overrightarrow{a}_{1} = \hat{i} + 2\hat{j} + 3\hat{k}, \overrightarrow{b}_{1} = \hat{i} - 3\hat{j} + 2\hat{k}$$\overrightarrow{a}_{2} = 4\hat{i} + 5\hat{j} + 6\hat{k}, \overrightarrow{b}_{2} = 2\hat{i} + 3\hat{j} + \hat{k}$
$\therefore \overrightarrow{a}_{2} - \overrightarrow{a}_{1} = 3\hat{i} + 3\hat{j} +3\hat{k}$
$\overrightarrow{b}_{1} \times\overrightarrow{b}_{2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{vmatrix} = -9\hat{i} + 3\hat{j} + 9\hat{k} $
$\therefore \begin{vmatrix} \overrightarrow{b}_{1}\times&\overrightarrow{b}_{2}\end{vmatrix} = {\sqrt{9^{2} + 3^{2} +9^{2}}} = \sqrt{171} $
$\therefore \text{Shorted distance d is given by}$
$d=\Bigg| \frac{\big(\overrightarrow{b_{1}}\times\overrightarrow{b_{1}}\big).{\big(\overrightarrow{a_{2}} - \overrightarrow{a_{1}}\big)}}{\big|\overrightarrow{b_{1}}\times\overrightarrow{b_{1}}\big|}\Bigg|$
$= \bigg|\frac{-27 + 9 + 27}{\sqrt{171}}\bigg|=\frac{9}{\sqrt{171}} =\frac{3}{\sqrt{19}}$
View full question & answer→Question 365 Marks
Find the angle between the lines whose direction cosines are given by the equations $l + m + n = 0, l^2 + m^2- n^2 = 0.$
AnswerEliminating $n$ from both the equations, we have $l^2 + m^2 - (l - m)^2 = 0$
$\Rightarrow l^2 + m^2 - l^2 - m^2 + 2ml = 0 $
$\Rightarrow 2lm = 0 $
$\Rightarrow lm = 0$
$\Rightarrow (-m - n)m = 0 [\because\text{l}=-\text{m}-\text{n}]$
$\Rightarrow m = -n$
$\Rightarrow m = 10$
$\Rightarrow l = 0, l = -n$
Thus, Dr’s two lines are proportional to $0, -n, n$ and $-n, 0, -1$ i.e., $0, -1, 1$ and $-1, 0, 1.$
So, the vector parallel to these given lines are $\vec{\text{a}}=-\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=-\hat{\text{i}}+\hat{\text{k}}$
Now, $\cos\theta=\frac{\vec{\text{a}}\vec{\text{b}}}{|\vec{\text{a}}||\vec{\text{b}}|}$
$\Rightarrow\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt{2}}$
$\Rightarrow\cos\theta=\frac{1}{2}$
$\therefore\theta=\frac{\pi}{3}\Big[\because\cos\frac{\pi}{3}=\frac{1}{2}\Big]$
View full question & answer→Question 375 Marks
Find the direction cosines of the line $\frac{4-\text{x}}{2}=\frac{\text{y}}{6}=\frac{1-\text{z}}{3}.$ Also, reduce it to vector form
AnswerThe cartesian equation of the given line is
$\frac{4-\text{x}}{2}=\frac{\text{y}}{6}=\frac{1-\text{z}}{3}$ It can be re-written as
$\frac{\text{x}-4}{-2}=\frac{\text{y}-0}{6}=\frac{\text{z}-1}{-3}$
This shows that the given line passes through the point 4, 0, 1 and its direction ratios are proportional to -2, 6, -3.
So, its direction cosines are
$\frac{-2}{\sqrt{(-2)^2+6^2+(-3)^2}},\frac{6}{\sqrt{(-2)^2+6^2+(-3)^2}},\frac{-3}{\sqrt{(-2)^2+6^2+(-3)^2}}$
$=\frac{-2}{7},\frac{6}7{},\frac{-3}{7}$
Thus, the given line passes through the points having position vector $\vec{\text{a}}=4\hat{\text{i}}+\hat{\text{k}}$ and is parallel to the vector $\vec{\text{b}}=-2\hat{\text{i}}+6\hat{\text{j}}-3\hat{\text{k}}$
We know that the vector equation of a line passing through a point with position vector a and parallel to the vector b is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
Here,
$\vec{\text{a}}=4\hat{\text{i}}+\hat{\text{k}};\vec{\text{b}}=-2\hat{\text{i}}+6\hat{\text{j}}-3\hat{\text{k}}$
Vector equation of the required line is
$\vec{\text{r}}=\big(4\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}\big)+\lambda\big(-2\hat{\text{i}}+6\hat{\text{j}}-3\hat{\text{k}}\big)$
Here, $\lambda$ is a parameter.
View full question & answer→Question 385 Marks
Find the vector and cartesian equations of the line passing through $(1, 2, 3)$ and parallel to the planes $\vec{\text{r}}\cdot(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=5$ and $\vec{\text{r}}\cdot(3\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})=6$
AnswerWe know that equation of a line passing through $(x_1, y_1, z_1)$ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{{\text{y}}-{\text{y}_1}}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}\ ...(\text{i})$
Here, required line is passing through $(1, 2, 3),$ is given by, $[$Using $(i)]$
$\frac{\text{x}-1}{\text{a}_1}=\frac{{\text{y}}-2}{\text{b}_1}=\frac{\text{z}-3}{\text{c}_1}\ ...(\text{ii})$
We know that, line $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{{\text{y}}-{\text{y}_1}}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1}$ is parallel to plane $a_2x + b_2y + c_2z + d_2 = 0$ if
$a_1a_2 + b_1b_2 + c_1c_{2 }= 0 ....(iii)$
Given, line $(ii)$ is parallel to plane $x - y + 2z = 5$
So, $a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a_1)(1) + (b_1)(-1) + (c_1)(2) = 0$
$a_1 + b_1 + 2c_1 = 0 ....(iv)$
Also, given line $(ii)$ is parallel to plane $3x + y + z = 6$
So, $a_1a_2 + b_1b_2 + c_1c_{2 }= 0$
$(a_1)(3) + (b_1)(1) + (c_1)(1) = 0$
$3a_1 + b_1 + c_1 = 0 ....(v)$
Solving $(iv)$ and $(v)$ by cross$-$multiplication,
$\frac{\text{a}_1}{(-1)(1)-(1)(2)}=\frac{\text{b}_1}{(3)(2)-(1)(1)}=\frac{\text{c}_1}{(1)(1)-(3)(-1)}$
$\frac{\text{a}_1}{-1-2}=\frac{\text{b}_1}{6-1}=\frac{\text{c}_1}{1+3}$
$\frac{\text{a}_1}{-3}=\frac{\text{b}_1}{5}=\frac{\text{c}_1}{4}=\lambda(\text{ say})$
$\text{a}_1=-3\lambda,\text{b}_1=5\lambda,\text{c}_1=4\lambda$
Put $a_1, b_1, c_1$ in equation $(ii),$
$\frac{\text{x}-1}{-3\lambda}=\frac{\text{y}-2}{5\lambda}=\frac{\text{z}-3}{4\lambda}$
Multiplying by $\lambda,$
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{5}=\frac{\text{z}-3}{4}$
Equation of required line is,
$\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{5}=\frac{\text{z}-3}{4}$
The vector equation of the line is
$\vec{\text{r}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+\lambda(-3\hat{\text{i}}+5\hat{\text{j}}+4\hat{\text{k}})$
View full question & answer→Question 395 Marks
Show that the lines $\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}$ and 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4 intersect. Find the equation of the plane in which they lie and also their of intersection.
AnswerThe equation of the given line is
$\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}$
The coordinates of any point on this line are of the form
$\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2}=\lambda$
$\Rightarrow\text{x}=3\lambda-4,\text{ y}=5\lambda-6,\text{ z}=-2\lambda+1$
So, the coordinates of the point on the given line are $(3\lambda-4,5\lambda-6,-2\lambda+1).$ Since this point lies on the plane
$3\text{x}-2\text{y}+\text{z}+5=0$
$3(3\lambda-4)-2(5\lambda-6)+(-2\lambda+1)+5=0$
$\Rightarrow9\lambda-12-10\lambda+12-2\lambda+1+5=0$
$\Rightarrow-3\lambda+6=0$
$\Rightarrow\lambda=2$
So, the coordinates of the point are
$(3\lambda-4,5\lambda-6,-2\lambda+1)$
$=\big(3(2)-4,5(2)-6,-2(2)+1\big)$
$=(2,4,-3)$
Substituting this point in another plane equation 2x + 3y + 4z - 4 = 0, we get
2(2) + 3(4) + 4(-3) - 4 = 0
⇒ 4 + 12 - 12 - 4 = 0
⇒ 0 = 0
So, the point (2, 4, -3) lies on another plane too. So, this point pf intersection of the lines.
Finding the plane equation
Let the direction ratios be proportional to a, b, c.
Since the plane contains the line $\frac{\text{x}+4}{3}=\frac{\text{y}+6}{5}=\frac{\text{z}-1}{-2},$ it must pass through the point (-4, -6, 1) and is parallel to this line.
So, the equation of plane is
a(x + 4) + b(y + 6) + c(z - 1) = 0 ....(i)
and 3a + 5b - 2c = 0 ....(ii)
Since the given plane contains the planes 3x - 2y + z + 5 = 0 = 2x + 3y + 4z - 4,
3a - 2b + c = 0 ...(iii)
2a + 3b + 4z = 0 ....(iv)
Solving (iii) and (iv) using cross-multiplication, we get
$\frac{\text{a}}{-11}=\frac{\text{b}}{-10}=\frac{\text{c}}{13}\ ....(\text{v})$
Using (i), (ii) and (v), the equation of plane is
$\begin{vmatrix}\text{x}+4&\text{y}+6&\text{z}-1\\3&-5&-2\\11&10&-13 \end{vmatrix}=0$
⇒ -45(x + 4) + 17(y + 6) - 25(z - 1) = 0
⇒ 45(x + 4) - 17(y + 6) + 25(z - 1) = 0
⇒ 45x - 17y + 25z + 53 = 0
View full question & answer→Question 405 Marks
Find the equation of the line passing through the points $(1, -1, 1)$ and perpendicular to the lines joining the points $(4, 3, 2), (1, -1, 0)$ and $(1, 2, -1) (2, 1, 1).$
AnswerWe know that equation of a line passing through $(x_1, y_{1, }z_1)$ and direction ratios as $a, b, c$ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}\dots(1)$
So, equation of a line passing through $(1, -1, 1)$ is
$\frac{\text{x}-1}{\text{a}}=\frac{\text{y}+1}{\text{b}}=\frac{\text{z}-1}{\text{c}}\dots(2)$
Now, Directions ratios of the line joining $A(4, 3, 2)$ and $B(1, -1, 0)$
$=(1 - 4), (-1 -3), (0 - 2)$
$\Rightarrow$ Direction ratios of line $AB = -3, -4, -2$
and, Direction ratios of the line joining $C(1, 2, -1)$ and $D(2, 1, 1)$
$=(2 - 1), (1 - 2), (1 + 1)$
$\Rightarrow$ Diraction ratios of line $CD =1, -1, 2$
Given that, line $AB$ is perpendicular to line $(2),$ so
$a_1a_{2 }+ b_1b_{2 }+ c_1c_{2 }= 0$
$(a)(-3) + (b)(-4) + (c)(-2) = 0$
$-3a + 4b - 2c = 0$
$3a + 4b + 2c = 0 .....(3)$
and, line $CD$ is also perpendicular to line $(2),$ so
$a_{1 }a_{2 }+ b_{1 }b_{2 }+ c_{1 }c_{2 }= 0$
$(a)(1) + (b)(-1) + (c)(2) = 0$
$a - b + 2c = 0 .....(4)$
Solving equation $(3)$ and $(4)$ using cross multiplication,
$\frac{\text{a}}{(4)(2)-(-1)(2)}=\frac{\text{b}}{(1)(2)-(3)(2)}=\frac{\text{c}}{(3)(-1)-(4)(1)}$
$\Rightarrow\frac{\text{a}}{8+2}=\frac{\text{b}}{2-6}=\frac{\text{c}}{-3-4}$
$\Rightarrow\frac{\text{a}}{10}=\frac{\text{b}}{-4}=\frac{\text{c}}{-7}=\lambda ($say$)$
$\Rightarrow\text{a}=10\lambda,\text{b}=-4\lambda,\text{c}=-7\lambda$
View full question & answer→Question 415 Marks
Find the equation of the lines joining the following pairs of vertices and then find the shortest distance between the lines(1) (0, 0, 0) and (1, 0, 2)
(2) (1, 3, 0) and (0, 3, 0)
AnswerEquation of line passing throught (0, 0, 0) and (1, 0, 2) is given by $\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$\vec{\text{r}}=\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)+\lambda\big((1-0)\hat{\text{i}}+(0-0)\hat{\text{j}}+(2-0)\hat{\text{k}}\big)$
$\vec{\text{r}}=\big(0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{k}}\big)\dots(1)$
Equation of another line passing through (1, 3, 0) and (0, 3, 0) is
$\vec{\text{r}}=\big(\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}\big)+\mu\big((0-1)\hat{\text{i}}+(3-3)\hat{\text{j}}+(0-0)\hat{\text{k}}\big)$
$\vec{\text{r}}=\big(\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}\big)\dots(2)$
From equation (1) and (2)
$\vec{\text{a}}_1=\big(0.\hat{\text{i}}+0.\hat{\text{j}}+0.\hat{\text{k}}\big),\vec{\text{b}}_1=\big(\hat{\text{i}}+2\hat{\text{k}}\big)$
$\vec{\text{a}}_2=\big(\hat{\text{i}}+3\hat{\text{j}}+0\hat{\text{k}}\big),\vec{\text{b}}_2=-\hat{\text{i}}$
we know that, shortest distance between the lines $\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$ and $\vec{\text{r}}=\vec{\text{a}}_2+\lambda\vec{\text{b}}_2$ is given by
$\text{S.D.}=\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\dots(3)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(\hat{\text{i}}+3\hat{\text{j}}+0.\hat{\text{k}}\big)-\big(0.\hat{\text{i}}+0.\hat{\text{j}}+0.\hat{\text{k}}\big)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)=\big(\hat{\text{i}}+3\hat{\text{j}}\big)$
$\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&0&2\\-1&0&0 \end{vmatrix}$
$=\hat{\text{i}}(0-0)-\hat{\text{j}}(0+2)+\hat{\text{k}}(-2)$
$\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=-2\hat{\text{j}}$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(\hat{\text{i}}+3\hat{\text{j}}\big)\big(-2\hat{\text{j}}\big)$
$=(1)(0)+(3)(-2)$
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=-6$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-2)^2}$
$\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=2$
Using $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)$ and $\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|$ in equation (1) to get shortest distance between the lines, so
$\text{S.D.}=\Big|\frac{-6}{2}\Big|$
$\text{S.D}=3\text{ units}$
View full question & answer→Question 425 Marks
Find the equation of the line passing through the points $(-1, 2, 1)$ and parallel to the line $\frac{2\text{x}-1}{4}=\frac{3\text{y}+5}{4}=\frac{2-\text{z}}{3}.$
AnswerWe know that, equation of a line passing through $(x_1, y_1, z_1)$ and direction ratios are $a, b, c$ is given by
$\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
Here, $(x_1, y_1, z_1) =(-1, 2, 1)$
and required line is parallel to the given line
$\frac{2\text{x}-1}{4}=\frac{3\text{y}+5}{2}=\frac{2-\text{z}}{3}$
$\Rightarrow\frac{\text{x}-\frac{1}{2}}{2}=\frac{\text{y}+\frac{5}{3}}{\frac{2}{3}}=\frac{\text{z}-2}{-3}$
$\Rightarrow $ Diraction ratios of the required line are proportional to $2,\frac{2}{3},-3$
$\Rightarrow\text{a}=2\lambda,\text{b}=\frac{2}{3}\lambda,\text{c}=-3\lambda$
So, required equation of the line using eqoation $(1),$
$\frac{\text{x}+1}{2\lambda}=\frac{\text{y}-2}{\frac{2}{3}\lambda}=\frac{\text{z}-1}{-3\lambda}$
$\frac{\text{x}+1}{2}=\frac{\text{y}-2}{\frac{2}{3}}=\frac{\text{z}-1}{-3}$
View full question & answer→Question 435 Marks
Find the foot of perpendicular from the point (2, 3, 4) to the line $\frac{4-\text{x}}{2}=\frac{\text{y}}{6}=\frac{1-\text{z}}{3}.$ Also, find the perpendicular distance from the given point to the line.
AnswerLet foot of the perpendicular from P(2, 3, 4) is $\theta$ on the line $\frac{4-\text{x}}{2}=\frac{\text{y}}{6}=\frac{1-\text{z}}{3},$ so
Equation of given line is,
$\frac{4-\text{x}}{2}=\frac{\text{y}}{6}=\frac{1-\text{z}}{3}$
$\frac{\text{x}-4}{-2}=\frac{\text{y}}{6}=\frac{\text{z}-1}{-3}=\lambda$ (say)
Coordinate of Q $=(-2\lambda+4,6\lambda,-3\lambda+1)$
Direction ratios of PQ $=(-2\lambda+4-2),(6\lambda-3),(-3\lambda+1-4)$
$=(-2\lambda+2),(6\lambda-3),(-3\lambda-3)$
Line PQ is perpendicular to given line, so
$\text{a}_1\text{a}_2+\text{b}_1\text{b}_2+\text{c}_1\text{c}_2=0$
$(-2)(-2\lambda+2)+(6)(6\lambda-3)+(-3)(-3\lambda-3)=0$
$4\lambda-4+36\lambda-18+9\lambda+9=0$
$49\lambda-13=0$
$\lambda=\frac{13}{49}$
Coordinate of Q $=(-2\lambda+4,6\lambda,-3\lambda+1)$
$=\Big(-2\Big(\frac{13}{49}\Big)+4,6\Big(\frac{13}{49}\Big),-3\Big(\frac{13}{49}\Big)+1\Big)$
$=\Big(\frac{-26+196}{49},\frac{78}{49},\frac{-39+49}{49}\Big)$
Coordinate of Q $=\Big(\frac{170}{49},\frac{78}{49},\frac{10}{49}\Big)$
$\text{PQ}=\sqrt{(\text{x}_1-\text{x}_2)^2+(\text{y}_1-\text{y}_2)^2+(\text{z}_1-\text{z}_2)^2}$
$=\sqrt{\Big(\frac{170}{49}-2\Big)^2+\Big(\frac{78}{49}-3\Big)^2+\Big(\frac{10}{49}-4\Big)^2}$
$=\sqrt{\frac{72}{49}^2+\Big(\frac{69}{49}\Big)^2+\Big(-\frac{168}{49}\Big)^2}$
$=\sqrt{\frac{5184+4761+34596}{2401}}$
$=\sqrt{\frac{44541}{2401}}$
$=\sqrt{\frac{909}{49}}$
$=\frac{3\sqrt{101}}{49}$
Perpendicular distance from (2, 3, 4) to given line is $\frac{3\sqrt{101}}{49}$ units.
View full question & answer→Question 445 Marks
Find the equation of line passing through the point $A(0, 6, -9)$ and $B(-3, -6, 3).$ If $D$ is the foot of perpendicular drawn from a point $C(7, 4, -1)$ on the line $AB,$ then find the coordiantes of the point $D$ and the equation of line $CD.$
AnswerEquation of line $AB$ is
$\frac{\text{x}-\text{x}_1}{\text{x}_2-\text{x}_1}=\frac{\text{y}-\text{y}_1}{\text{y}_2-\text{y}_1}=\frac{\text{z}-\text{z}_1}{\text{z}_2-\text{z}_1}$
$\Rightarrow\frac{\text{x}-0}{-3-0}=\frac{\text{y}-6}{-6-6}=\frac{\text{z}+9}{3+9}$
$\Rightarrow\frac{\text{x}}{-3}=\frac{\text{y}-6}{-12}=\frac{\text{z}+9}{12}=\lambda$ (say)
Coordinate of point D $=\big(-3\lambda,-12\lambda+6,12\lambda-9\big)$
Direction ratios of CD $=(-3\lambda-7),(-12\lambda+6-4),(12\lambda-9+1)$
$=(-3\lambda-7),(-12\lambda+2),(12\lambda-8)$
Line CD is perpendicular to line $AB,$ so
$a_1a_{2 }+ b_1b_{2 }+ c_1c_{2 }= 0$
$\Rightarrow(-3)(-3\lambda-7)+(-12)(-12\lambda+2)+(12)(12\lambda-8)=0$
$\Rightarrow9\lambda+21+144\lambda-24+144\lambda-96=0$
$\Rightarrow297\lambda-99=0$
$\Rightarrow\lambda=\frac{1}{3}$
Coordinate of D $=\big(-3\lambda,-12\lambda+6,12\lambda-9\big)$
$=\Big(-3\big(\frac{1}{3}\big),-12\big(\frac{1}{3}\big)+6,12\big(\frac{1}{3}\big)-9\Big)$
Coordinate of $D = (-1, 2, -5)$
Equation of $CD$ is,
$\frac{\text{x}-\text{x}_1}{\text{x}_2-\text{x}_1}=\frac{\text{y}-\text{y}_1}{\text{y}_2-\text{y}_1}=\frac{\text{z}-\text{z}_1}{\text{z}_2-\text{z}_1}$
$\Rightarrow\frac{\text{x}-7}{-1-7}=\frac{\text{y}-4}{2-4}=\frac{\text{z}+1}{-5+1}$
$\Rightarrow\frac{\text{x}-7}{-8}=\frac{\text{y}-4}{-2}=\frac{\text{z}+1}{-4}$
or $\frac{\text{x}-7}{4}=\frac{\text{y}-4}{1}=\frac{\text{z}+1}{2}$
View full question & answer→Question 455 Marks
Find the foot of the perpendicular drawn from the point $\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$ to the line $\vec{\text{r}}=\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big).$ Also, find the length of the perpendicylar
AnswerLet $\angle$ be the foot of the perpendicular drawn from the point $\text{p}\big(\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)$ to the line $\vec{\text{r}}=\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big).$ Let the position vector $\angle$ be $\vec{\text{r}}=\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)$ $=\lambda\hat{\text{i}}+(1+2\lambda)\hat{\text{j}}+(2+3\lambda)\hat{\text{k}}\dots(1)$
Now, $\overrightarrow{\text{PL}}=$ Position vector of L - Position vector of P $\Rightarrow\overrightarrow{\text{PL}}=\Big\{\lambda\hat{\text{i}}+(1+2\lambda)\hat{\text{j}}+(2+3\lambda)\hat{\text{k}}\Big\}-\big(\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}\big)$ $\Rightarrow\overrightarrow{\text{PL}}=(\lambda-1)\hat{\text{i}}+(2\lambda-5)\hat{\text{j}}+(3\lambda-1)\hat{\text{k}}\dots(2)$ Since $\overrightarrow{\text{PL}}$ is perpandicular to the given line, which is parallel to $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}},$ we have $\overrightarrow{\text{PL}}.\vec{\text{b}}=0$ $\Rightarrow\Big\{(\lambda-1)\hat{\text{i}}+(2\lambda-5)\hat{\text{j}}+(3\lambda-1)\hat{\text{k}}\Big\}.\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)=0$ $\Rightarrow1(\lambda-1)+2(2\lambda-5)+3(3\lambda-1)=0$ $\Rightarrow\lambda=1$ Substituting $\lambda=1$ in (1), we get the position vector of $\angle$ as $\hat{\text{i}}+3\hat{\text{j1}}+5\hat{\text{k}}.$ Substituting $\lambda=1$ in (2), we get $\overrightarrow{\text{PL}}=-3\hat{\text{j}}+2\hat{\text{k}}$ $=\sqrt{(-3)^2+2^2}$ $=\sqrt{13}$ Hence, the length of the perpendicular from point P on PL is $\sqrt{13}\text{ units}.$ View full question & answer→Question 465 Marks
Find the angle between the follwing pairs of lines$: \vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$ and $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}-\mu\big(2\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}\big)$
Answer$\vec{\text{r}}=\big(4\hat{\text{i}}-\hat{\text{j}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$ and $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}-\mu\big(2\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}\big)$
Let $b_1$ and $b_2$ be vectors parallel to the given lines.
Now,
$\vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
$\vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}$
If $\theta$ is the angle between the given lines, then
$\cos\theta=\frac{\vec{\text{b}}_1.\vec{\text{b}}_2}{\big|\vec{\text{b}}_1\big|\big|\vec{\text{b}}_2\big|}$
$=\frac{\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big).\big(2\hat{\text{i}}+4\hat{\text{j}}-4\hat{\text{k}}\big)}{\sqrt{1^2+2^2+(-2)^2}\sqrt{2^2+4^2+(-4)^2}}$
$=\frac{2+8+8}{3\times6}$
$=1$
$\Rightarrow\theta=0^{\circ}$
View full question & answer→Question 475 Marks
Find the value of so $\lambda$ that the lines$\frac{1 - x}{3} = \frac{7y - 14}{2\lambda} = \frac{5z - 10}{11} \text{and} \frac{7 - 7x}{3\lambda} = \frac{y - 5}{1}= \frac{6 - z}{5}$
are perpendicular to each other.
AnswerGetting direction rations of two lines as$(a_{1}, b_{1}, c_{1},) = \bigg( -3, \frac{2\lambda}{7}, \frac{11}{5}\bigg) \text{and} (a_{2}, b_{2}, c_{2},) = \bigg(-\frac{3\lambda}{7},1, -5\bigg)$
Two lines are perpendicular then $a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0$
$\Rightarrow \frac{9\lambda}{7} + \frac{2\lambda}{7} - 11 = 0 \Rightarrow \frac{11\lambda}{7} = 11 \Rightarrow \lambda = 7$
View full question & answer→Question 485 Marks
Find the foot of the perpendicular from (1, 2, -3) on the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{-2}=\frac{\text{z}}{-1}.$
AnswerLet L be the foot of the perpendicular drawn from the point P(1, 2, -3) to the given line. The coordinates of a general point on the line $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{-2}=\frac{\text{z}}{-1}=\lambda$ are given by $\frac{\text{x}+1}{2}=\frac{\text{y}-3}{-2}=\frac{\text{z}}{-1}=\lambda$ $\Rightarrow\text{x}=2\lambda-1$ $\text{y}=-2\lambda+3$ $\text{z}=-\lambda$ Let the coordinates of L be $2\lambda-1,-2\lambda+3,-\lambda.$
The direction ratios of PL are proportional to $2\lambda-1-1,-2\lambda+3-2,-\lambda+3,$ i.e. $2\lambda-2,-2\lambda+1,-\lambda+3.$ The direction ratios of the given line are proportional to 2, -2, -1, but PL is perpendicular to the given line. $\therefore22\lambda-2-2-2\lambda+1-1-\lambda+3=0\Rightarrow\lambda=1$ Substituting $\lambda=1$ in $2\lambda-1,-2\lambda+3,-\lambda,$ we get the coordinates of L as 1, 1, -1. View full question & answer→Question 495 Marks
Find the vector equation of a line passing through the point with position vector $\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$ and parallel to the line joining the points with position vectors $\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}$ and $2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}.$ Also, find the cartesian equivalent of this equation.
AnswerWe know that, equation of a line passing through $\vec{\text{a}}$ and parallel to vector $\vec{\text{b}}$ is,
$\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\dots(1)$
Here, $\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
and, $\vec{\text{b}}=$ line joining $\big(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)$ and $\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)$
$=\big(2\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}+4\hat{\text{k}}\big)$
$=2\hat{\text{i}}-\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+2\hat{\text{k}}-4\hat{\text{k}}$
$=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Equation of the line is
$\vec{\text{r}}=\big(\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\big)$
For cortesion form of equation put $\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}},$
$\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}=(1+\lambda)\hat{\text{i}}+(-2+2\lambda)\hat{\text{j}}+(-3-2\lambda)\hat{\text{k}}$
Equating coeffcients of $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ so
$\text{x}=1+\lambda,\text{y}=-2+2\lambda,\text{z}=-3-2\lambda$
$\Rightarrow\frac{\text{x}-1}{1}=\lambda,\frac{\text{y}+2}{2}=\lambda,\frac{\text{z}+3}{-2}=\lambda$
So, $\frac{\text{x}-1}{1}=\frac{\text{y}+2}{2}=\frac{\text{z}+3}{-2}$
View full question & answer→Question 505 Marks
Find the Vector and Cartesian equations of the line passing through the point (1, 2, – 4) and perpendicular to the two lines $\frac{\text{x - 8}}{3} =\frac{\text{y + 19}}{-16}=\frac{\text{z - 10}}{7}\text{ and }\frac{\text{x - 15}}{3}= \frac{\text{y - 29}}{8}=\frac{\text{z - 5}}{-5}$.
AnswerEquations of the line can be written as $\frac{\text{x - 1}}{a} =\frac{\text{y - 2}}{b}=\frac{\text{z + 4}}{c}$ since the line is perpendicular to two given lines
$\therefore$ 3a – 16b + 7c = 0 and 3a + 8b – 5c = 0
$\therefore$ $\frac{\text{a}}{24} =\frac{\text{b}}{36}=\frac{\text{c}}{{72}}$
Hence the equations of the line are
$\frac{\text{x - 1}}{24}= \frac{\text{y - 2}}{36}=\frac{\text{z + 4}}{72}\text{ OR }\frac{\text{x - 1}}{2}= \frac{\text{y - 2}}{3}=\frac{\text{z + 4}}{6}$
and the vector form is
$\overrightarrow{r}=(\hat{\text{i}}+\hat{\text{2j}}-\hat{\text{4k}})+\lambda(\hat{\text{2i}}+\hat{\text{3j}}+\hat{\text{6k}})$
View full question & answer→