Question 1515 Marks
Find the vector and Cartesian equations of the line through the point (1, 2, – 4) and perpendicular to the two lines.
$\overrightarrow{\text{r}} = (8\hat{\text{i}} - 19\hat{\text{j}} + 10\hat{\text{k}})+\lambda(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}$ and $\overrightarrow{\text{r}} = (15\hat{\text{i}} - 29\hat{\text{j}} + 5\hat{\text{k}})+\mu(3\hat{\text{i}} - 8\hat{\text{j}} + 5\hat{\text{k})}.$
AnswerVector equation of the required line is
$\overrightarrow{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - 4\hat{\text{k}})+\mu[(3\hat{\text{i}} - 16\hat{\text{j}} + 7\hat{\text{k})}\times(3\hat{\text{i}} + 8\hat{\text{j}} - 5\hat{\text{k})]}$
$\Rightarrow\overrightarrow{\text{r}} = (\hat{\text{i}} + 2\hat{\text{j}} - 4\hat{\text{k}})+\lambda[(2\hat{\text{i}} + 3\hat{\text{j}} + 6\hat{\text{k})]}$
in cartesian form, $\frac{\text{x - 1}}{2} = \frac{\text{y - 2}}{3} = \frac{\text{z + 4}}{6}$
View full question & answer→Question 1525 Marks
Find the equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2x + 3y + 4z = 5$ which is perpendicular to the plane $x - y + z = 0.$
AnswerThe equation of plane through the intersection of planes,
$x + y + z = 1$ and $2x + 3y + 4z = 5,$ is
$(\text{x + y + z}-1)+\lambda(2\text{x}+3\text{y}+4\text{z}-5)=0$
$\Rightarrow\ \ (2\lambda+1)\text{x}+(3\lambda+1)\text{y}+(4\lambda+1)\text{z}-(5\lambda+1)=0\ \ ....(1)$
The direction ratios, $a_1, b_1, c_1,$ are $1, -1,$ and $1.$
Since the planes are perpendicular,
$a_1a_2 + b_1b_2 + c_1c_2 = 0$
$\Rightarrow\ \ (2\lambda+1)-(3\lambda+1)+(4\lambda+1)=0$
$\Rightarrow\ \ 3\lambda+1=0$
$\Rightarrow\ \ \lambda=-\frac{1}{3}$
Substituting $\lambda=-\frac{1}{3}$ in equation $(1),$ we obtain
$\frac{1}{3}\text{x}-\frac{1}{3}\text{z}+\frac{2}{3}=0$
$\Rightarrow x - z + 2 = 0$
This is the required equation of the plane.
View full question & answer→Question 1535 Marks
Find the value of $\lambda$ such that the line $\frac{\text{x}-2}{6}=\frac{\text{y}-1}{\lambda}=\frac{\text{z}+5}{-4}$ is perpendicular to the plane 3x - y - 2z = 7.
AnswerHere, given mid line $\frac{\text{x}-2}{6}=\frac{\text{y}-1}{\lambda}=\frac{\text{z}+5}{-4}$ is parpendicular to plane 3x - y - 2z = 7 so, normal vector of plane is parallel to line so,
Direction ratios of normal to plane are proparional to the direction ratios of line.
Here,
$\frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}$
cross multiplying the last two
$-2\lambda=4$
$\lambda=\frac{4}{-2}$
$\lambda=-2$
View full question & answer→Question 1545 Marks
Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: $\frac{\text{x}-8}{3}=\frac{\text{y}+19}{-16}=\frac{\text{z}-10}{7}\ \text{and}\ \frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}.$
AnswerGiven: A point on the Required line is A(1, 2, -4)
$\therefore\ \text{Position vector of point A is }\vec{\text{a}}=(1,2,-4)=\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}$
Also given equations of two lines
$\frac{\text{x}-8}{3}=\frac{\text{y}+19}{-16}=\frac{\text{z}-10}{7}\ \text{and}\ \frac{\text{x}-15}{3}=\frac{\text{y}-29}{8}=\frac{\text{z}-5}{-5}$
$\therefore$ Direction ratios of given two lines are
$\vec{\text{b}_1}=3\hat{\text{i}}-16\hat{\text{j}}+7\hat{\text{k}}\ \text{and}\ \vec{\text{b}_2}=3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}$
$\vec{\text{b}}=\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\3&-16&7\\3&8&-5\end{vmatrix}$
Expanding along first row,
$=\hat{\text{i}}(80-56)-\hat{\text{j}}(-15-21)+\hat{\text{k}}(24+48)=24\hat{\text{i}}+36\hat{\text{j}}+72\hat{\text{k}}$
$\Rightarrow\ \vec{\text{b}}=12\Big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\Big)$
$\therefore$ Equation of the required line is $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}$
$\Rightarrow \vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\Big)+\lambda(12)\Big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\Big)$
Again replacing $12\lambda\ \text{by}\ \lambda$
$\Rightarrow \vec{\text{r}}=\Big(\hat{\text{i}}+2\hat{\text{j}}-4\hat{\text{k}}\Big)+\lambda\Big(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}\Big).$
View full question & answer→Question 1555 Marks
Find the acute angle between the lines whose direction ratios are proportional to 2 : 3 : 6 and 1 : 2 : 2.
AnswerThe vectors, represented by these are
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}}$
and $\vec{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Let, $\theta$ be the angle between the lines,
then,
$\cos\theta=\frac{\vec{\text{a}}\times\vec{\text{b}}}{\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|}$
$=\frac{(2\hat{\text{i}}+3\hat{\text{j}}+6\hat{\text{k}})(\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})}{\sqrt{(2)^2+(3)^2+(6)^2}\sqrt{(1)^2+(2)^2+(2)^2}}$
$=\frac{(2)(1)+(3)(2)+(6)(2)}{\sqrt{4+9+36}\sqrt{1+4+4}}$
$=\frac{2+6+12}{\sqrt{49}\sqrt{9}}$
$=\frac{20}{7\times3}$
$\cos\theta=\frac{20}{21}$
$\theta=\cos^{-1}\Big(\frac{20}{21}\Big)$
Angle between the lines $=\cos^{-1}\Big(\frac{20}{21}\Big)$.
View full question & answer→Question 1565 Marks
Find the coordinates of the point where the line through $(5, 1, 6)$ and $(3, 4, 1)$ crosses the $YZ-$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 $YZ-$plane
i.e., $x = 0 .......(ii)$
Putting $x = 0$ in eq. $(i),$ we get
$\frac{-5}{-2}=\frac{\text{y}-1}{3}=\frac{\text{z}-6}{-5}$
$\Rightarrow\ \ \frac{\text{y}-1}{3}=\frac{5}{2}\ \text{and}\ \frac{\text{z}-6}{-5}=\frac{5}{2}$
$\Rightarrow 2y - 2 = 15$ and $2z - 12 = -25$
$\Rightarrow 2y = 17$ and $2z = -13$
$\Rightarrow\ \ \ \text{y}=\frac{17}{2}\ \text{and}\ \text{z}=\frac{-13}{2}$
Thus, required point is $\text{P}\Big(0,\ \frac{17}{2},\ \frac{-13}{2}\Big).$
View full question & answer→Question 1575 Marks
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.
AnswerEquation of line through A(3, 4, 1) and B(5, 1, 6)
$\frac{\text{x - 3}}{2} = \frac{\text{y - 4}}{-3} = \frac{\text{z - 1}}{5} = \text{k (say)}$
General point on the line:
$\text{x = 2k + 3, y = -3k + 4, z = 5k + 1}$
line crosses xz plane i.e. $\text{y = 0 if -3k + 4 = 0}$
$\therefore\text{k} = \frac{4}{3}$
Co-ordinate of required point $\bigg(\frac{17}{3}, 0, \frac{23}{3}\bigg)$
Angle, which line makes with xz plane:
$\sin\theta = \bigg|\frac{2(0) + (-3) (1) + 5 (0)}{\sqrt{4 + 9 + 25}\sqrt{1}}\bigg| = \frac{3}{\sqrt{38}}\Rightarrow \theta = \sin^{-1}\bigg(\frac{3}{\sqrt{38}}\bigg)$
View full question & answer→Question 1585 Marks
Find the vector equation of the plane passing through the points P(2, 5, -3), Q(-2, -3, 5) and R(5, 3, -3).
Answer
The required plane passes through the point P(2, 5, -3), whose position vector is $\vec{\text{a}}=2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}}$ and is normal to the vector $\vec{\text{n}}$ given by
$\overrightarrow{\text{PQ}}=\overrightarrow{\text{OQ}}-\overrightarrow{\text{OP}}$
$=(-2\hat{\text{i}}-3\hat{\text{j}}+5\hat{\text{k}})-(2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}})$
$=-4\hat{\text{i}}-8\hat{\text{j}}+8\hat{\text{k}}$
$\overrightarrow{\text{PR}}=\overrightarrow{\text{OR}}-\overrightarrow{\text{OP}}$
$=(5\hat{\text{i}}+3\hat{\text{j}}-3\hat{\text{k}})-(2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}})$
$=3\hat{\text{i}}-2\hat{\text{j}}-0\hat{\text{k}}$
$\vec{\text{n}}=\overrightarrow{\text{PQ}}\times\overrightarrow{\text{PR}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-4&-8&8\\3&-2&0\end{vmatrix}$
$=16\hat{\text{i}}+24\hat{\text{j}}+32\hat{\text{k}}$
The vector equation of the required plane is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(16\hat{\text{i}}+24\hat{\text{j}}+32\hat{\text{k}})=(2\hat{\text{i}}+5\hat{\text{j}}-3\hat{\text{k}})\cdot(16\hat{\text{i}}+24\hat{\text{j}}+32\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[8(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\Big]=32+120-96$
$\Rightarrow\vec{\text{r}}\cdot\Big[8(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})\Big]=56$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})=7$ View full question & answer→Question 1595 Marks
Show that the vectors $\overrightarrow{\text{a}},\overrightarrow{\text{b}} \text{and}{\overrightarrow{\text{c}}}$ are coplanar $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}}+\overrightarrow{\text{c}}\text{and} \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar.
AnswerGiven that $\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}$ are coplanar
$\therefore[\overrightarrow{\text{a}} +\overrightarrow{\text{b}}, \overrightarrow{\text{b}} + \overrightarrow{\text{c}}, \overrightarrow{\text{c}} + \overrightarrow{\text{a}}] = 0$
$\text{i.e} \overrightarrow{\text{(a}} +\overrightarrow{\text{b})}, \big\{\overrightarrow{\text{(b}} + \overrightarrow{\text{c})} \times\overrightarrow{\text{(c}} + \overrightarrow{\text{a})}\big\} = 0$
$\overrightarrow{\text{(a}} +\overrightarrow{\text{b})}. \big\{\overrightarrow{\text{(b}} \times \overrightarrow{\text{c}}+ \overrightarrow{\text{b}} \times\overrightarrow{\text{a}}\times\overrightarrow{\text{c}} \times \overrightarrow{\text{a})}\big\} = 0$
$\Rightarrow\overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{a}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{a}}(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})}+\overrightarrow{\text{b}}. (\overrightarrow{\text{b}}\times\overrightarrow{\text{c})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{b}}\times\overrightarrow{\text{a})} + \overrightarrow{\text{b}}.(\overrightarrow{\text{c}}\times\overrightarrow{\text{a})} = 0$
$\Rightarrow 2 [\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}] = \text{0 or} [\overrightarrow{\text{a}},\overrightarrow{\text{b}}, \overrightarrow{\text{c] }} = 0$
$\Rightarrow\overrightarrow{\text{a}}, \overrightarrow{\text{b}}, \overrightarrow{\text{c}}\text{are coplanar}.$
View full question & answer→Question 1605 Marks
Find the direction cosines of the sides of the triangle whose vertices are (3, 5, –4), (–1, 1, 2) and (–5, –5, –2).
AnswerDirection ratios of the line joining A and B are -1 - 3, 1 - 5, 2 - (-4)
$\Rightarrow-4,\ -4,\ 6\ \ \ \ \ \ \ \ [\because\ \text{x}_2-\text{x}_1,\ \text{y}_2-\text{y}_1]$
$\therefore$ Direction cosines of line AB are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\Rightarrow\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-4)^2+(6)^2}},\ \frac{6}{\sqrt{(-4)^2+(-4)^2+(6)^2}}$
$\Rightarrow\ \frac{-4}{\sqrt{16+16+36}},\ \frac{-4}{\sqrt{16+16+36}},\ \frac{6}{\sqrt{16+16+36}}$
$\Rightarrow\ \frac{-4}{\sqrt{68}},\ \frac{-4}{\sqrt{68}},\ \frac{6}{\sqrt{68}}\ \ \Rightarrow\ \frac{-4}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}},\ \frac{6}{2\sqrt{17}}$
$\Rightarrow\ \frac{-2}{\sqrt{17}},\ \frac{-2}{\sqrt{17}},\ \frac{3}{\sqrt{17}}$
Now Direction ratios of the line joining B and C are -5 - (-1), -5 - 1, -2 - 2 = -4, -6, -4
$\therefore$ Direction cosines of line BC are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\therefore$ Direction cosines of line BC are
$\Rightarrow\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-6}{\sqrt{(-4)^2+(-6)^2+(-4)^2}},\ \frac{-4}{\sqrt{(-4)^2+(-6)^2+(-4)^2}}$
$\Rightarrow\ \frac{-4}{\sqrt{16+36+16}},\ \frac{-6}{\sqrt{16+36+16}},\ \frac{-4}{\sqrt{16+36+16}}$
$\Rightarrow\ \frac{-4}{\sqrt{68}},\ \frac{-6}{\sqrt{68}},\ \frac{-4}{\sqrt{68}}\ \ \Rightarrow\ \frac{-4}{2\sqrt{17}},\ \frac{-6}{2\sqrt{17}},\ \frac{-4}{2\sqrt{17}}$
$\Rightarrow\ \frac{-2}{\sqrt{17}},\ \frac{-3}{\sqrt{17}},\ \frac{-2}{\sqrt{17}}$
Direction ratios of the line joining C and A are 3 - (-5), 5 - (-5), -4 - (-2) = 8, 10, -2
$\therefore$ Direction cosines of line CA are
$\frac{\text{a}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{b}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}},\ \frac{\text{c}}{\sqrt{\text{a}^2+\text{b}^2+\text{c}^2}}$
$\Rightarrow\ \frac{8}{\sqrt{(8)^2+(10)^2+(-2)^2}},\ \frac{10}{\sqrt{(8)^2+(10)^2+(-2)^2}},\ \frac{-2}{\sqrt{(8)^2+(10)^2+(-2)^2}}$
$\Rightarrow\ \frac{8}{\sqrt{64+100+4}},\ \frac{10}{\sqrt{64+100+4}},\ \frac{-2}{\sqrt{64+100+4}}$
$\Rightarrow\ \frac{8}{\sqrt{168}},\ \frac{10}{\sqrt{168}},\ \frac{-2}{\sqrt{168}}\ \ \Rightarrow\ \frac{8}{2\sqrt{42}},\ \frac{10}{2\sqrt{42}},\ \frac{-2}{2\sqrt{42}}$
$\Rightarrow\ \frac{4}{\sqrt{42}},\ \frac{5}{\sqrt{42}},\ \frac{-1}{\sqrt{42}}.$
View full question & answer→Question 1615 Marks
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XZ plane. Also find the angle which this line makes with the XZ plane.
AnswerEquation of line through A(3, 4, 1) and B(5, 1, 6)
$\frac{\text{x - 3}}{2} = \frac{\text{y - 4}}{-3} = \frac{\text{z - 1}}{5} = \text{k (say)}$
General point on the line:
$\text{x = 2k + 3, y = -3k + 4, z = 5k + 1}$
line crosses xz plane i.e. $\text{y = 0 if -3k + 4 = 0}$
$\therefore\text{k} = \frac{4}{3}$
Co-ordinate of required point $\bigg(\frac{17}{3}, 0, \frac{23}{3}\bigg)$
Angle, which line makes with xz plane:
$\sin\theta = \bigg|\frac{2(0) + (-3) (1) + 5 (0)}{\sqrt{4 + 9 + 25}\sqrt{1}}\bigg| = \frac{3}{\sqrt{38}}\Rightarrow \theta = \sin^{-1}\bigg(\frac{3}{\sqrt{38}}\bigg)$
View full question & answer→Question 1625 Marks
Find the value of p, so that the lines $l_1$ :$\frac{1-\text{x}}{3}=\frac{7\text{y}-\text{14}}{\text{p}}=\frac{\text{z}-\text{3}}{2}$ and $l_2$: $\frac{7-\text{7x}}{3\text{p}}=\frac{\text{y}-\text{5}}{1}=\frac{6-\text{z}}{5}$are perpendicular to each other. Also find the equations of a line passing through a point (3, 2,– 4) and parallel to line $l_1$.
AnswerGiven lines can be written as
$l_1$:$\frac{1-\text{x}}{-3}=\frac{\text{y}-\text{2}}{\text{p}/7}=\frac{\text{z}-\text{3}}{2}$ ; $l_2$: $\frac{\text{x}-\text{1}}{-3\text{p}/7}=\frac{\text{y}-\text{5}}{1}=\frac{\text{z}-6}{-5}$
since the lines are perpendicular
$\therefore\ (-3)\big(-\frac{3\text{p}}{7}\big)+\big(\frac{\text{p}}{7}\big)(1)+(2)(-5)=0$
$\Rightarrow$ p = 7
Equation of line passing through (3, 2, – 4) and parallel to $l_1$ is
$\frac{\text{x}-3}{-3}=\frac{\text{y}-\text{2}}{\text{1}}=\frac{\text{z}+\text{4}}{2}$
View full question & answer→Question 1635 Marks
The equation of tangent at (2, 3) on the curve $y^{2} = \text{ax}^{3} + \text{b is y = 4x - 5}. $ Find the value of a and b.
Answer$\text{y}^{2} = \text{ax}^{3} + \text{b} \Rightarrow\text{2y}\frac{\text{dy}}{\text{dx}} = \text{3ax}^{2}\therefore \frac{\text{dy}}{\text{dx}} = \frac{\text{3a}}{2} \frac{\text{x}^{2}}{\text{y}}$
Slope of tangent at (2, 3) $\frac{\text{dy}}{\text{dx}}\bigg]_{(2, 3)} =\frac{\text{3a}}{2}. \frac{4}{3} = \text{2a}$
Comparing with slope of tangent $\text{y = 4x - 5, we get, 2a = 4}\therefore$
Also (2, 3) lies on the curve $\therefore 9 = \text{8a + b, put a = 2, we get b = -7}$
View full question & answer→Question 1645 Marks
Find the vector and the cartesian equations of the lines that passes through the origin and $(5, -2, 3).$
Answer$\vec{\text{a}}=$ Position vector of a point here $O ($say$)$ on the line $(0,\ 0,\ 0)=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\vec{0}$
$\vec{\text{b}}=$ A vector along the line $=\overrightarrow{\text{OA}}=$ Position vector of a point $A -$ Position vector of $ O=(5,-2,\ 3)-(0,\ 0,\ 0)=(5,-2,\ 3)=5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
$\therefore$ Vector equation of the line is $\Big(\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}}\Big)$
$\Rightarrow\ \ \vec{\text{r}}=\vec{0}+\lambda\Big(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)$
$\Rightarrow\ \ \vec{\text{r}}=\lambda\Big(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)$
Now Cartesian equation of the line Direction ratios of line $OA$ are $5 - 0, -2 - 0, 3 - 0 = 5, -2, 3$ And a point on the line is $O(0, 0, 0) = (x_1, y_1, z_1)$
$\therefore$ Cartesian equation of the line $=\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$
$=\frac{\text{x}-0}{5}=\frac{\text{y}-0}{-2}=\frac{\text{z}-0}{3}=\frac{\text{x}}{5}=\frac{\text{y}}{-2}=\frac{\text{z}}{3}$
Remark: In the solution of the above question we can also take:
$\vec{\text{a}}=$ Position vector of point $A = (5, -2, 3) $
$=5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ for vector form and point A as $(x_1, y_1, z_1) = (5, -2, 3)$ for Cartesian form.
Then the equation of the line in vector form is $\vec{\text{r}}=5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}+\lambda\Big(5\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)$
And equation of line in Cartesian form is $\frac{\text{x}-5}{5}=\frac{\text{y}+2}{-2}=\frac{\text{z}-3}{3}$
View full question & answer→Question 1655 Marks
Find the equation of the plane passing through the point $(-1, 3, 2)$ and perpendicular to each of the planes $x + 2y + 3z = 5$ and $3x + 3y + z = 0.$
AnswerSince equation of any plane through the point $(-1, 3, 2)$ is $a(x - x_1) + b(y - y_1) + c(z - z_1) = 0$
$\therefore a(x + 1) + b(y - 3) + c(z - 2) .....(i)$
$\Rightarrow ax + a + by - 3b + cz - 2c = 0$
$\Rightarrow ax + by + cz = -a + 3b + 2c$
This required plane is perpendicular to the olane $x + 2y + 3z = 5(a_1a_2 + b_1b_2 + c_1c_2 = 0)$
$\therefore$ Products of coefficients $\Rightarrow\ \text{a}(1)+\text{b}(2)+\text{c}(3)=0\ \ \ ....(\text{ii})$
Again, the required plane is perpendicular to the plane $3x + 3y + z = 0$
$\therefore$ Products of coefficients $\Rightarrow\ \text{a}(3)+\text{b}(3)+\text{c}(1)=0\ \ \ ....(\text{iii})$
Solving eq. $(ii)$ and $(iii),$ we get
$\frac{\text{a}}{2-9}=\frac{\text{b}}{9-1}=\frac{\text{c}}{3-6}\ \ \ \Rightarrow\ \ \frac{\text{a}}{-7}=\frac{\text{b}}{8}=\frac{\text{c}}{-3}$
Putting these values of $a, b, c$ in eq. $(i),$ we get
$-7(x + 1) + 8(y - 3) - 3(z - 2) = 0$
$\Rightarrow -7x - 7 + 8y - 24 - 3z + 6 = 0$
$\Rightarrow -7x + 8y - 3z - 25 = 0$
$\Rightarrow 7x - 8y + 3z + 25 = 0.$
View full question & answer→Question 1665 Marks
Find the value of $\lambda,$ so that the lines $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$ are at right angles. Also, find whether the lines are intersecting or not.
AnswerGiven lines are $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$
Converting them into standard form,
we have $=\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{\Big(\frac{\lambda}{7}\Big)}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\Big(\frac{-3\lambda}{7}\Big)}=\frac{\text{y}-5}{1}=\frac{\text{z}-6}{-5}$
View full question & answer→Question 1675 Marks
Find the value of $\lambda$ for which the lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{\lambda^2}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{\lambda^2}=\frac{\text{z}-1}{2}$ are coplanar.
AnswerThe 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 $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
The given lines $\frac{\text{x}-1}{1}=\frac{\text{y}-2}{2}=\frac{\text{z}+3}{\lambda^2}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-2}{\lambda^2}=\frac{\text{z}-1}{2}$ are coplanar.
$\therefore\ \begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}3-1&2-2&1-(-3)\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix}2&0&4\\1&2&\lambda^2\\1&\lambda^2&2\end{vmatrix}=0$
$\Rightarrow2(4-\lambda^4)-0+4(\lambda^4-2)=0$
$\Rightarrow-2\lambda^4+4\lambda^2=0$
$\Rightarrow\lambda^2(\lambda^2-2)=0$
$\Rightarrow\lambda^2=0\text{ or }\lambda^2-2=0$
$\Rightarrow\lambda=0\text{ or }\lambda=\pm\sqrt{2}$
Thus, the values of $\lambda$ are $0,-\sqrt{2}$ and $\sqrt{2}$
View full question & answer→Question 1685 Marks
Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.
AnswerEquations of line AB are
$\frac{\text{x-3}}{\text{2}}=\frac{\text{y-4}}{-3}=\frac{\text{z-1}}{5}=\lambda\ ......\text{(i)}$
the general point on (i) is
$2\lambda + 3,\ – 3\lambda + 4,\ 5\lambda + 1$
the line (i) crosses XY-plane, then z = 0
$\lambda=-\frac{1}{5}$
hence point is $\bigg[2\Big(\frac{-1}{5}\Big)+3,\ -3\Big(\frac{-1}{5}\Big)+4,\ 5\Big(\frac{-1}{5}\Big)+1\bigg]$
$=\Big[\frac{13}{5},\ \frac{23}{5},\ 0\Big]$
View full question & answer→Question 1695 Marks
Find the shortest distance between the lines
$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$ $\text{and}\ \vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big).$
Answer$\vec{\text{r}}=\Big(4\hat{\text{i}}-\hat{\text{j}}\Big)+\lambda\Big(\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}\Big)$
$\vec{\text{a}}_1=4\hat{\text{i}}-\hat{\text{j}}\ \ \ \vec{\text{b}}_1=\hat{\text{i}}+2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{r}}=\Big(\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}\Big)+\mu\Big(2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}\Big)$
$\vec{\text{a}}_2=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}_2=2\hat{\text{i}}+4\hat{\text{j}}-5\hat{\text{k}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\cdot\big(\vec{\text{b}}_2\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1-\vec{\text{b}}_2\big|}\end{vmatrix}$
$\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}} &\hat{\text{j}}&\hat{\text{k}}\\1 & 2&-3\\2&4&-5 \end{vmatrix}$
$=\hat{\text{i}}(-10+12)-\hat{\text{j}}(-5+6)+\hat{\text{k}}(4-4)$
$=2\hat{\text{i}}-\hat{\text{j}}$
$\text{S.D.}=\begin{vmatrix}\frac{\big(-3\hat{\text{i}}+0\hat{\text{j}}+2\hat{\text{k}}\big)\cdot\big(2\hat{\text{i}}-\hat{\text{j}}\big)}{\sqrt{4+1}}\end{vmatrix}$
$=\begin{vmatrix}\frac{-6}{\sqrt{5}}\end{vmatrix}=\frac{6}{\sqrt{5}}$
View full question & answer→Question 1705 Marks
$\overrightarrow{\text{n}}$ is a vector of magnitude $\sqrt{3}$ and is equally inclined to an acute angle with the coordinate axes. Find the vector and cartesian form of the equation of a plane which passes through (2, 1, -1) and is normal to $\overrightarrow{\text{n}}$
AnswerHere, it is given that $\vec{\text{n}}=\sqrt{3}$ and $\vec{\text{n}}$ makes equal angle with coordinate axes.
Let, $\vec{\text{n}}$ has direction cosine as l. m and n and it makes angle of $\alpha,\beta$ and $\gamma$ with the coordinate axes, so
Here, $\alpha=\beta=\gamma$
$\Rightarrow\cos\alpha=\cos\beta=\cos\gamma$
$\Rightarrow\text{l}=\text{m}=\text{n}=\text{p}(\text{say})$
We know that,
$\text{l}^2+\text{m}^2+\text{n}^2=1$
$\text{p}^2+\text{p}^2+\text{p}^2=1$
$3\text{p}^2=1$
$\text{p}^2=\frac{1}{3}$
$\text{p}=\pm\frac{1}{\sqrt{3}}$
So,
$\text{l}=\pm\frac{1}{\sqrt{3}}$
$\cos\alpha=\pm\frac{1}{\sqrt{3}}$
Now, $\alpha=\cos^{-1}\Big(-\frac{1}{\sqrt{3}}\Big)$
It gives, $\alpha$ is an obtuse angle so, neglect it.
Again, $\alpha=\cos^{-1}\Big(-\frac{1}{\sqrt{3}}\Big)$
It gives, $\alpha$ is an acute angle, so
$\cos\alpha=\frac{1}{\sqrt{3}}$
$\therefore\ \text{l}=\text{m}=\text{n}=\frac{1}{\sqrt{3}}$
So,
$\vec{\text{n}}=|\vec{\text{n}}|(\text{l}\hat{\text{i}}+\text{m}\hat{\text{j}}+\text{n}\hat{\text{k}})$
$=\sqrt{3}\Big(\frac{1}{\sqrt{3}}\hat{\text{i}}+\frac{1}{\sqrt{3}}\hat{\text{j}}+\frac{1}{\sqrt{3}}\hat{\text{k}}\Big)$
$\vec{\text{n}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
And, $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$
We know that, vector equation of a plane passing through the point $\vec{\text{a}}$ and perpendicular to the vector $\vec{\text{n}}$ is given by,
$(\vec{\text{r}}-\vec{\text{a}}){\vec{\text{n}}}=0$
$\big[\vec{\text{r}}-(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})\big]\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-[(2)(1)+(1)(1)+(-1)(1)]=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-[2+1-1]=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})-2=0$
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2$
Put, $\vec{\text{r}}=(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})$
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2$
$(\text{x})(1)+(\text{y})(1)+(\text{z})(1)=2$
$\text{x}+\text{y}+\text{z}=2$
So, vector and cartesian equation of the plane is,
$\vec{\text{r}}\cdot(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}})=2$
$\text{x}+\text{y}+\text{z}=2$
View full question & answer→Question 1715 Marks
Find the equation of the plane through the points (2, 1, -1) and (-1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
AnswerThe equation of the plane passing through (2, 1, -1) is
a(x - 2) + b(y - 1) + c(z + 1) = 0 ......(i)
Since, this passes through (-1, 3, 4).
$\therefore$ a(-1 - 2) + b(3 - 1) + c(4 + 1) = 0
⇒ -3a + 2b + 5c = 0 ......(ii)
Since, the plane (i) is perpendicular to the plane x - 2y + 4z = 10.
$\therefore$ 1× a - 2 × b + 4 × c = 0
⇒ a - 2b + 4c = 0 ......(iii)
On solving equations (ii) and (iii), by cross multiplication method, we get
$\frac{\text{a}}{8+10}=\frac{\text{-b}}{-17}=\frac{\text{c}}{4}=\lambda$
$\Rightarrow\text{a}=18\lambda,\text{b}=17\lambda,\text{c}=4\lambda$
From Eq. (i),
$18\lambda(\text{x}-2)+17\lambda(\text{y}-1)+4\lambda(\text{z}-1)=0$
$\Rightarrow18\text{x}-36+17\text{y}-17-4\text{z}+4=0$
$\therefore18\text{x}+17\text{y}+4\text{z}-49$
View full question & answer→Question 1725 Marks
Find the equation of the perpendicular drawn from the point P(-1, 3, 2) to the line $\vec{\text{r}}=\big(2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big).$ Also, find the coordinates of the foot of the perpendicular from P.
AnswerLet Q be the perpendicular drow from $\text{p}\big(\hat{-\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$ on the line
$\vec{\text{r}}=\big(2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
Let the position vector of Q be
$\big (2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)$
$\big(2\lambda\big)\hat{\text{i}}+\big(2+\lambda\big)\hat{\text{j}}+\big(3+3\lambda\big)\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=$ position vector of Q-position vector of P
$\big\{\big(2\lambda\big)\hat{\text{i}}+\big(2+\lambda\big)\hat{\text{j}}+\big(3+3\lambda\big)\hat{\text{k}}\big\}-\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)$
$=(2\lambda+1)\hat{\text{i}}+(2\lambda-3)\hat{\text{j}}+(3+3\lambda-2)\hat{\text{k}}$
$\overrightarrow{\text{PQ}}=(2\lambda+1)\hat{\text{i}}+(\lambda-1)\hat{\text{j}}+(3\lambda+1)\hat{\text{k}}$
Since, $\overrightarrow{\text{PQ}}$ is perpendicular to given line, so
$\big\{(2\lambda+1)\hat{\text{i}}+(\lambda-1)\hat{\text{j}}+(3\lambda+1)\hat{\text{k}}\big\}\big(2\hat{\text{i}}+\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$(2\lambda+1)(2)+(\lambda-1)(1)+(3\lambda+1)3=0$
$4\lambda+2+\lambda-1+9\lambda+3=0$
$14\lambda+4=0$
$\lambda=-\frac{4}{14}$
$\lambda=-\frac{2}{7}$
Position vector of Q $=(2\lambda)\hat{\text{i}}+(2+\lambda)\hat{\text{j}}+(3+3\lambda)\hat{\text{k}}$
$=2\Big(-\frac{2}{7}\Big)\hat{\text{i}}+\Big(2-\frac{2}{7}\Big)\hat{\text{j}}+\Big(3+3\Big(-\frac{2}{7}\Big)\Big)\hat{\text{k}}$
$=-\frac{4}{7}\hat{\text{i}}+\frac{12}{7}\hat{\text{j}}+\frac{15}{7}\hat{\text{k}}$
Coordinates of foot of the perpendicular $=\Big(-\frac{4}{7},\frac{12}{7},\frac{15}{7}\Big)$
Equation of PQ is
$\vec{\text{r}}=\vec{\text{a}}+\lambda\big(\vec{\text{b}}-\vec{\text{a}}\big)$
$\Rightarrow\vec{\text{r}}=\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)+\lambda\Big(\Big(-\frac{4}{7}\hat{\text{i}}+\frac{12}{7}\hat{\text{j}}+\frac{15}{7}\hat{\text{k}}\Big)-\big(-\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big)\Big)$
View full question & answer→Question 1735 Marks
Find the shortest distance between the lines whose vector equations are:
$\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}\ \text{and}$
$\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}-(2\text{s}+1)\hat{\text{k}}$
AnswerEquation of first line is $\vec{\text{r}}=(1-\text{t})\hat{\text{i}}+(\text{t}-2)\hat{\text{j}}+(3-2\text{t})\hat{\text{k}}$
$\hat{\text{i}}-\text{t}\hat{\text{i}}+\text{t}\hat{\text{j}}-2\hat{\text{j}}+3\hat{\text{k}}-2\text{t}\hat{\text{k}}$
$=\Big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)+\text{t}\Big(-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{a}_1}+\text{t}\vec{\text{b}_1},$
$\vec{\text{a}_1}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}},\ \ \ \vec{\text{b}_1}=-\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
Equation of second line is $\vec{\text{r}}=(\text{s}+1)\hat{\text{i}}+(2\text{s}-1)\hat{\text{j}}+(2\text{s}+1)\hat{\text{k}}$
$\text{s}\hat{\text{i}}+\hat{\text{i}}+2\text{s}\hat{\text{j}}-\hat{\text{j}}-2\text{s}\hat{\text{k}}-\hat{\text{k}}$
$=\Big(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)+\text{s}\Big(\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}\Big)$
Comparing this equation with $\vec{\text{a}_2}+\text{s}\vec{\text{b}_2},$
$\vec{\text{a}_2}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}},\ \ \ \vec{\text{b}_2}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}$
Now shortest distance $(\text{d})=\frac{\Big|\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)\Big|}{\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|}\ \ \ \ ...(\text{i})$
$\vec{\text{a}_2}-\vec{\text{a}_1}=\Big(\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\Big)-\Big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\Big)=\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{b}_1}\times\vec{\text{b}_2}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\-1&1&-2\\1&2&-2\end{vmatrix}$
$=(-2+4)\hat{\text{i}}-(2+2)\hat{\text{j}}+(-2-1)\hat{\text{k}}=2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}$
$\Big|\vec{\text{b}_1}\times\vec{\text{b}_2}\Big|=\sqrt{(2)^2+(-4)^2+(-3)^2}=\sqrt{29}$
$\Big(\vec{\text{a}_2}-\vec{\text{a}_1}\Big).\Big(\vec{\text{b}_1}\times\vec{\text{b}_2}\Big)=\Big(\hat{\text{j}}-4\hat{\text{k}}\Big).\Big(2\hat{\text{i}}-4\hat{\text{j}}-3\hat{\text{k}}\Big)$
=0 × 2 + 1 × (-4) + (-4)(-3) = 8
Putting these values in eq.(i),
Shortest distance $(\text{d})=\frac{|8|}{\sqrt{29}}=\frac{8}{\sqrt{29}}.$
View full question & answer→Question 1745 Marks
Find the shortest distance between the following pairs of lines whose cartesian equation are:
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\text{z}$ and $\frac{\text{x}+2}{3}=\frac{\text{y}-2}{1};\text{z}=2$
AnswerThe equation of the given are
$\frac{\text{x}-1}{2}=\frac{\text{y}+1}{3}=\frac{\text{z}-0}{1}\dots(1)$
$\frac{\text{x}+1}{3}=\frac{\text{y}-2}{1}=\frac{\text{z}-2}{0}\dots(2)$
Since line (1) passes through the point (1, -1, 0) and has direction ratios proportional to 2, 3, 1, its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_1+\lambda\vec{\text{b}}_1$
Here,
$\vec{\text{a}}_1=\hat{\text{i}}-\hat{\text{j}}+0\hat{\text{k}}$
$\vec{\text{b}}_1=2\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$
Also, line (2) passes through the point (-1, 2, 2) and has diraction ratios proportional to 3, 1, 0.
Its vector equation is
$\vec{\text{r}}=\vec{\text{a}}_2+\mu\vec{\text{b}}_2$
Here,
$\vec{\text{a}}_2=-\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{b}}_2=3\hat{\text{i}}+\hat{\text{j}}+0\hat{\text{k}}$
Now,
$\vec{\text{a}}_2-\vec{\text{a}}_1=-2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}$
and $\vec{\text{b}}_1\times\vec{\text{b}}_2=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&3&1\\3&1&0\end{vmatrix}$
$=-\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|=\sqrt{(-1)^2+3^2(-7)^2}$
$=\sqrt{1+9+49}$
$=\sqrt{59}$
and $\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)=\big(-2\hat{\text{i}}+3\hat{\text{j}}+2\hat{\text{k}}\big).\big(-\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}\big)$
$=2+9-14$
$=-3$
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
$\text{d}=\Bigg|\frac{\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big).\big(\vec{\text{b}}_1\times\vec{\text{b}}_2\big)}{\big|\vec{\text{b}}_1\times\vec{\text{b}}_2\big|}\Bigg|$
$=\Big|\frac{-3}{\sqrt{59}}\Big|$
$=\frac{3}{\sqrt{59}}$
View full question & answer→Question 1755 Marks
Find the distance between the point (-1, -5, -10) and the point of intersection of the line $\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{12}$ and the plane $x - y + z = 5.$
AnswerAny point on the line $\frac{\text{x} - 2}{3} = \frac{\text{y} + 1}{4} = \frac{\text{z} - 2}{12} = \text{is} (3\lambda + 2, 4\lambda - 1, 12\lambda + 2)$
If this is the point of intersection with plane $\text{x - y + z = 5}$
$\text{then 3} \lambda + 2 - 4\lambda + 1 + 12\lambda + 2 - 5 = 0 \Rightarrow \lambda = 0$
$\therefore$ Point of intersection is (2, -1, 2)
Required distance = $\sqrt{(2 + 1)^{2} +(-1 + 5)^{2} + (2 + 10)^{2} } = 13$
View full question & answer→Question 1765 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 1775 Marks
Show that the lines $\frac{5 - x }{-4} =\frac{\text{y}- 7}{4} = \frac{\text{z} + 3}{-5}\text{ and } \frac{{x} - 8}{7} =\frac{2\text{y} - 8}{2} =\frac{\text{z} - 5 }{3}$ are coplanar.
AnswerEquations of lines are:
$\frac{x - 5}{4} =\frac{\text{y}- 7}{4} = \frac{\text{z} + 3}{-5};\ \frac{{x} - 8}{7} =\frac{\text{y} - 4}{1} =\frac{\text{z} - 5 }{3}$
Here, $x_1 = 5, y_1 = 7, z_1 = – 3 ; x_2 = 8, y_2 = 4, z_2 = 5$
$a_1 = 4, b_1 = 4, c_1 = – 5 ; a_2 = 7, b_2 = 1, c_2 = 3$
$ \begin{vmatrix} \text{x}_2-\text{x}_1 & \text{y}_2-\text{y}_1 & \text{z}_2-\text{z}_1 \\ \text{a}_1 & \text{b}_1 &\text{c}_1 \\ \text{a}_2 & \text{b}_2 & \text{c}_2 \end{vmatrix}$$= \begin{vmatrix} 3 & -3 & 8 \\ 4 & 4 & -5 \\ 7 & 1 & 3 \end{vmatrix}= 3(17) + 3 (47) + 8 (– 24) = 0$
$\therefore$ lines are co-planar
View full question & answer→Question 1785 Marks
Find the equation of the plane which is parallel to 2x - 3y + z = 0 and which passes through (1, -1, 2).
AnswerGiven, equation of plane is,
2x - 3y + z = 0 ....(i)
We know that equation of a plane parallel the plane (i) is given by
$2\text{x}-3\text{y}+\text{z}+\lambda=0\ ....(\text{ii})$
Given that, plane (ii) is passing through the point (1, -1, 2) so it must satisfy the equation (ii),
$2(1)-3(-1)+(2)+\lambda=0$
$2+3+2+\lambda=0$
$7+\lambda=0$
$\lambda=-7$
Put the value of $\lambda$ in equation (ii),
2x - 3y + z - 7 = 0
So, equation of the required plane is,
2x - 3y + z = 7
View full question & answer→Question 1795 Marks
Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.
AnswerWe know to find the equation pf plane that bisects A(1, 2, 3) and B(3, 4, 5) perpendicularly
We know that, equation of plane passing through the point $\vec{\text{a}}$ and perpendicular to vector $\vec{\text{n}}$ is given by,
$(\vec{\text{r}}-\vec{\text{a}})\vec{\text{n}}=0\ ...(\text{i})$
Here, $\vec{\text{a}}=\text{mid-point of AB}$
$=\frac{\text{position vector of A}+\text{position vector of B}}{2}$
$=\frac{\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}+3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}}{2}$
$\vec{\text{a}}=\frac{4\hat{\text{i}}+6\hat{\text{j}}+8\hat{\text{k}}}{2}$
$\vec{\text{a}}=2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}}$
And, $\vec{\text{n}}=\overrightarrow{\text{AB}}$
= Position vector of B - Position vector of A
$=(3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}})-(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})$
$=3\hat{\text{i}}+4\hat{\text{j}}+5\hat{\text{k}}-\hat{\text{i}}-2\hat{\text{j}}-3\hat{\text{k}}$
$\vec{\text{n}}=2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$
Put, the value of $\vec{\text{a}}$ and $\vec{\text{n}}$ in equation (i),
$\vec{\text{r}}-(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-\big[(2\hat{\text{i}}+3\hat{\text{j}}+4\hat{\text{k}})(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})\big]=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-[(2)(2)+(3)(2)+(4)(2)]=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-[4+6+8]=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})-18=0$
$\vec{\text{r}}(2\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}})=18$
View full question & answer→Question 1805 Marks
If the line $\frac{\text{x}-3}{2}=\frac{\text{y}+2}{-1}=\frac{\text{z}+4}{3}$ lies in the plane $lx + my - z = 9,$ then find the value of $l^2 + m^2.$
AnswerThe line $\frac{\text{x}-\text{x}_1}{\text{a}}=\frac{\text{y}-\text{y}_1}{\text{b}}=\frac{\text{z}-\text{z}_1}{\text{c}}$ lies in the plane
$Ax + By + Cz + D = 0$
if $(i) Ax_1 + By_1 + Cz_1 + D = 0$ and $(ii)\ aA + bB + cC = 0$
It is given that the line $\frac{\text{x}-3}{2}=\frac{\text{y}+2}{-1}=\frac{\text{z}+4}{3}$ lies in the plane $lx + my - z = 9$
$\therefore l \times 3 + m \times (-2) - (-4) = 9$
$\Rightarrow 3l - 2m = 5 ....(i)$
Also,
$2 \times l + (-1) \times m + 3 \times (-1) = 0$
$\Rightarrow 2l - m = 3 ....(ii)$
Solving $(i)$ and $(ii)$ we get
$l = 1$ and $m = -1$
$\therefore l^2 + m^2 = 1^2+ (-1)^2 = 1 + 1 = 2$
Thus, the value of $l^2 + m^2$ is $ 2$
View full question & answer→Question 1815 Marks
The equation of tangent at (2, 3) on the curve $y^{2} = \text{ax}^{3} + \text{b is y = 4x - 5}. $ Find the value of a and b.
Answer$\text{y}^{2} = \text{ax}^{3} + \text{b} \Rightarrow\text{2y}\frac{\text{dy}}{\text{dx}} = \text{3ax}^{2}\therefore \frac{\text{dy}}{\text{dx}} = \frac{\text{3a}}{2} \frac{\text{x}^{2}}{\text{y}}$
Slope of tangent at (2, 3) $\frac{\text{dy}}{\text{dx}}\bigg]_{(2, 3)} =\frac{\text{3a}}{2}. \frac{4}{3} = \text{2a}$
Comparing with slope of tangent $\text{y = 4x - 5, we get, 2a = 4}\therefore$
Also (2, 3) lies on the curve $\therefore 9 = \text{8a + b, put a = 2, we get b = -7}$
View full question & answer→Question 1825 Marks
Find the vector equation (in scalar product form) of the plane containing the line of intersection of the planes x - 3y + 2z - 5 = 0 and 2x - y + 3z - 1 = 0 and passing through (1, -2, 3).
AnswerThe equation of the plane passing through the line of intersection of the given planes is
$\text{x}-3\text{y}+2\text{z}-5+\lambda(2\text{x}-\text{y}+3\text{z}-1)=0\ ...(\text{i})$
This passing through (1, -2, 3). So,
$1+6+6-5+\lambda(2+2+9-1)$
$\Rightarrow8+12\lambda=0$
$\Rightarrow\lambda=\frac{-2}{3}$
Substituting this in (i) we get
$\text{x}-3\text{y}+2\text{z}-5-\frac{2}{3}(2\text{x}-\text{y}+3\text{z}-1)=0$
$\Rightarrow-\text{x}-7\text{y}-13=0$
$\Rightarrow\text{x}+7\text{y}+13=0$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}+7\hat{\text{j}})+13=0,$ Which is the required vector equation of the plane.
View full question & answer→Question 1835 Marks
Find the equation of the plane through the point $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ and passing throught the line of intersection of the planes $\vec{\text{r}}\cdot(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})=0$ and $\vec{\text{r}}\cdot(\hat{\text{j}}+2\hat{\text{k}})=0.$
AnswerThe equation of the plane passing through the line of intersection of the given planes is,
$\vec{\text{i}}(\hat{\text{i}}+3\hat{\text{j}}-\hat{\text{k}})+\lambda\big(\vec{\text{r}}(\hat{\text{j}}+2\hat{\text{k}})\big)=0$
$\vec{\text{r}}\cdot\Big[\hat{\text{i}}+(3+\lambda)\hat{\text{j}}+(-1+2\lambda)\hat{\text{k}}\Big]=0\ ...(\text{i})$
This passes through $2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}$ So,
$(2\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}})\Big[\hat{\text{i}}+(3+\lambda)\hat{\text{j}}+(-1+2\lambda)\hat{\text{k}}\Big]=0$
$\Rightarrow2+3+\lambda+1-2\lambda=0$
$\Rightarrow\lambda=6$
Substituting this in (i), we get
$\vec{\text{r}}\cdot\Big[\hat{\text{i}}+(3+6)\hat{\text{j}}+(-1+12)\hat{\text{k}}\Big]=0$
$\Rightarrow\vec{\text{r}}(\hat{\text{i}}+9\hat{\text{j}}+11\hat{\text{k}})=0$
View full question & answer→Question 1845 Marks
The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k} \text{ and } 2\hat{i} + 2\hat{j} + 3\hat{k}.$ Find the two unit vectors parallel its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Answer$\text{let } \text{d}_{1} \& \text{ d}_{2}$ be the two diagonal vectors:
$\therefore \overrightarrow{\text{d}}_{1} = 4\hat{\text{i}} - 2\hat{\text{j}} - 2\hat{\text{k}}, \overrightarrow{\text{d}}_{2} = -6\hat{\text{j}} - 8\hat{\text{k}}$
$\text{or} \overrightarrow{\text{d}}_{2} = 6\hat{\text{j}} + 8\hat{\text{k}}$
Unit vectors parallel to the diagonals are:
$\overrightarrow{\text{d}}_{1} = \frac{2}{\sqrt{6}}\hat{\text{i}} - \frac{1}{\sqrt{6}}\hat{\text{j}} - \frac{1}{\sqrt{6}}\hat{\text{k}}$
$\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} - \frac{4}{5}\hat{\text{k}} \bigg(\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} + \frac{4}{5}\hat{\text{k}}\bigg)$
$\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & -2 & -2 \\ 0 & -6 & -8 \end{vmatrix} = 4\hat{\text{i}} + 32\hat{\text{j}} - 24\hat{\text{k}} $
Area of parallelogram $= \frac{1}{2}|\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2}| = \sqrt{404} \text{ or } 2\sqrt{101} \text{sq. units}$
View full question & answer→Question 1855 Marks
If the lines $\text{x}=5,\frac{\text{y}}{3-\alpha}=\frac{\text{z}}{-2}$ and $\text{x}=\alpha,\frac{\text{y}}{-1}=\frac{\text{z}}{2-\alpha}$ are coplanar, find the values of $\alpha.$
AnswerThe 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 $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2}$ are coplanar if
$\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}=0$
The given lines $\frac{\text{x}-5}{0}=\frac{\text{y}}{3-\alpha}=\frac{\text{z}}{-2}$ and $\frac{\text{x}-\alpha}{0}=\frac{\text{y}}{-1}=\frac{\text{z}}{2-\alpha}$ are coplanar.
$\therefore\ \begin{vmatrix} \alpha-5&0-0&0-0\\0&3-\alpha&-2\\0&-1&2-\alpha\end{vmatrix}=0$
$\Rightarrow\begin{vmatrix} \alpha-5&0-0&0-0\\0&3-\alpha&-2\\0&-1&2-\alpha\end{vmatrix}=0$
$\Rightarrow(\alpha-5)\Big[(3-\alpha)\times(2-\alpha)-2\Big]-0+0=0$
$\Rightarrow(\alpha-5)(\alpha-1)(\alpha-4)=0$
$\Rightarrow\alpha-1=0\text{ or }\alpha-4=0\text{ or }\alpha-5=0$
$\Rightarrow\alpha=1\text{ or }\alpha=4\text{ or }\alpha=5$
Thus, the values of $\alpha$ are 1, 4 and 5.
View full question & answer→Question 1865 Marks
$\text{Let } \vec{\text a} = \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}, \vec{\text{b}} = \hat{\text{i}} \text{ and } \vec{\text{c}} = \text{c}_{1} \hat{\text{i}} + \text{c}_{2} \hat{\text{j}} + \text{c}_{3} \hat{\text{k}}, \text{then}$
- Let $c_1 = 1$ and $c_2 = 2$, find $c_3$ which makes $\vec{\text{a}}, \vec{\text{b}} \text{ and }\vec{\text{c}} \text{ coplanar.}$
- If $c_2 = –1$ and $c_3 = 1,$ show that no value of $c_1$ can make $\vec{\text{a}}, \vec{\text{b}} \text{ and } \vec{\text{c}} \text{ coplanar}.$
Answer$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ \text{c}_{1} & \text{c}_{2} & \text{c}_{3} \end{vmatrix} = \text{c}_{2} - \text{c}_{3}$
- $\text{c}_{1} = 1, \text{ c}_{2} = 2$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 2 - \text{c}_{3}$
$\therefore \vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ are coplanar} [\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = 0 \Rightarrow \text{c}_{3} = 2$
- $\text{c}_{2} = -1, \text{c}_{3} = 1$
$[\vec{\text{a}} \text{ }\vec{\text{b}} \text{ } \vec{\text{c}}] = \text{c}_{2} - \text{c}_{3} = -2 \neq 0$
$\Rightarrow \text{No value of }\text{c}_{1} \text{can make }\vec{\text{a}}, \text{ }\vec{\text{b}}, \text{ } \vec{\text{c}} \text{ coplanar}.$ View full question & answer→Question 1875 Marks
Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, -4, -5) and B(2, -3, 1) intersects the plane 2x + y + z = 7.
AnswerThe equation of a line joining the points A(3, -4, -5) and B(2, -3, 1) is
$\frac{\text{x}-3}{2-3}=\frac{\text{y}+4}{-3+4}=\frac{\text{z}+5}{1+5}=\text{r}$
$\Rightarrow\text{x}=3-\text{r},\text{ y}=-4+\text{r},\text{ z}=-5+6\text{r}$
Substituting this into the given plane equation we get
$2(3-\text{r})+(-4+\text{r})+(-5+6\text{r})=7$
$\Rightarrow\text{r}=2$
$\Rightarrow\text{x}=1,\text{ y}=-2,\text{ z}=7$
Distance of (1, -2, 7) from (3, 4, 4) is
$=\sqrt{(3-1)^2+(4+2)^2+(4-7)^2}$
$=\sqrt{4+36+9}$
$=\sqrt{49}$
$=7$
View full question & answer→Question 1885 Marks
Find the cartesian form of the equations of the following planes.
$\vec{\text{r}}=(1+\text{s}+\text{t})\hat{\text{i}}+(2-\text{s}+\text{t})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
AnswerThe given equation of the plane is,
$\vec{\text{r}}=(1+\text{s}+\text{t})\hat{\text{i}}+(2-\text{s}+\text{t})\hat{\text{j}}+(3-2\text{s}+2\text{t})\hat{\text{k}}$
$\Rightarrow\vec{\text{r}}=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})+\text{s}(\hat{\text{i}}-\hat{\text{j}}-2\hat{\text{k}})+\text{t}(\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}})$
Normal vector, $\vec{\text{n}}=\vec{\text{b}}\times\vec{\text{c}}$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-1&-2\\1&1&2\end{vmatrix}$
$=0\hat{\text{i}}-4\hat{\text{j}}+2\hat{\text{k}}$
$=-4\hat{\text{j}}+2\hat{\text{k}}$
The vector equation of the plane in scalar product from is
$\vec{\text{r}}\cdot\vec{\text{n}}=\vec{\text{a}}\cdot\vec{\text{n}}$
$\Rightarrow\vec{\text{r}}\cdot(-4\hat{\text{j}}+2\hat{\text{k}})=(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}})(-4\hat{\text{j}}+2\hat{\text{k}})$
$\Rightarrow\vec{\text{r}}\cdot\Big[-2(2\hat{\text{j}}-\hat{\text{k}})\Big]=0-8+6$
$\Rightarrow\vec{\text{r}}\cdot\Big[-2(2\hat{\text{j}}-\hat{\text{k}})\Big]=-2$
$\Rightarrow\vec{\text{r}}\cdot(2\hat{\text{j}}-\hat{\text{k}})=1$
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
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})\cdot(2\hat{\text{j}}-\hat{\text{k}})=1$
$\Rightarrow2\text{y}-\text{z}=1$
View full question & answer→Question 1895 Marks
Find the angle between the vectors with direction ratios proportional to 1, -2, 1 and 4, 3, 2.
AnswerLet $\vec{\text{a}}$ be a vector with direction ratios 1, -2, 1.
$\Rightarrow\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}$
Let $\vec{\text{b}}$ be a vector with direction ratios 4, 3, 2.
$\Rightarrow\vec{\text{b}}=4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}$
Let $\theta$ be the angle between the given vectors.
Now,
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|}$
$=\frac{(\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}).(4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}})}{\big|\hat{\text{i}}-2\hat{\text{j}}+\hat{\text{k}}\big|\big|4\hat{\text{i}}-3\hat{\text{j}}+2\hat{\text{k}}\big|}$
$=\frac{4-6+2}{\sqrt{1+4+1}\sqrt{16+9+4}}$
$=\frac{0}{\sqrt{6}\sqrt{29}}$
$=0$
$\therefore\theta=\frac{\pi}{2}$
Thus, the angle between the given vectors measures $\frac{\pi}{2}$.
View full question & answer→Question 1905 Marks
Find the shortest distance between the following pairs of parallel lines whose equations are:$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$ and $\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
AnswerThe vector equation of the given lines are
$\vec{\text{r}}=\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big)+\lambda\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$\vec{\text{r}}=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)+\mu\big(-\hat{\text{i}}+\hat{\text{j}}-\hat{\text{k}}\big)$
$=\big(2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}\big)-\mu\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
These two lines pass through the points having position vectors $\vec{\text{a}}_1=\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}$ and $\vec{\text{a}}_2=2\hat{\text{i}}-\hat{\text{j}}-\hat{\text{k}}$ and are parallel to the vector $\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Now,
$\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}=\big(\hat{\text{i}}-3\hat{\text{j}}-4\hat{\text{k}}\big)\times\big(\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}\big)$
$=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\1&-3&-4\\1&-1&1 \end{vmatrix}$
$=-7\hat{\text{i}}-5\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|=\sqrt{(-7)^2+(-5)^2+2^2}$
$=\sqrt{49+25+4}$
$=\sqrt{78}$
The shortest distance between the two lines is given by
$\frac{\big|\big(\vec{\text{a}}_2-\vec{\text{a}}_1\big)\times\vec{\text{b}}\big|}{\big|\vec{\text{b}}\big|}=\frac{\sqrt{78}}{\sqrt{3}}=\sqrt{26}$
View full question & answer→Question 1915 Marks
Find the value of $\lambda,$ so that the lines $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$ are at right angles. Also, find whether the lines are intersecting or not.
AnswerGiven lines are $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$
Converting them into standard form,
we have $=\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{\Big(\frac{\lambda}{7}\Big)}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\Big(\frac{-3\lambda}{7}\Big)}=\frac{\text{y}-5}{1}=\frac{\text{z}-6}{-5}$
View full question & answer→Question 1925 Marks
Find the value of $\lambda,$ so that the lines $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$ are at right angles. Also, find whether the lines are intersecting or not.
AnswerGiven lines are $\frac{1-\text{x}}{3}=\frac{\text{7}\text{y}-14}{\lambda}=\frac{\text{z}-3}{2}$ and $=\frac{7-7\text{x}}{3\lambda}=\frac{\text{y}-5}{1}=\frac{6-\text{z}}{5}$
Converting them into standard form,
we have $=\frac{\text{x}-1}{-3}=\frac{\text{y}-2}{\Big(\frac{\lambda}{7}\Big)}=\frac{\text{z}-3}{2}$ and $\frac{\text{x}-1}{\Big(\frac{-3\lambda}{7}\Big)}=\frac{\text{y}-5}{1}=\frac{\text{z}-6}{-5}$
View full question & answer→Question 1935 Marks
Find the equation of the plane through the line of intersection of the planes $x + 2y + 3z + 4 = 0$ and $x - y + z + 3 = 0$ and passing through the origin.
AnswerWe know that, equation of a plane passing through the line of intersection of planes $a_1x + b_1y + c_1z + d_1 = 0$ and $a_2x + b_2y + c_2z + d_2 = 0$ is given by,
$(\text{a}_1\text{x}+\text{b}_1\text{y}+\text{c}_1\text{z}+\text{d}_1)+\lambda(\text{a}_2\text{x}+\text{b}_2\text{y}+\text{c}_2\text{z}+\text{d}_2)=0$
So, equation of plane passing through the line of intersection of planes $x + 2y + 3z + 4 = 0 $ and $x - y + z + 3 = 0$ is
$(\text{x}+2\text{y}+3\text{z}+4)+\lambda(\text{x}-\text{y}+\text{z}+3)=0$
$\text{x}(1+\lambda)+\text{y}(2-\lambda)+\text{z}(3+\lambda)+4+3\lambda=0\ ...(\text{i})$
Equation $(i)$ is passing through origin, so
$(0)(1+\lambda)+(0)(2-\lambda)+(0)(3+\lambda)+4+3(\lambda)=0$
$0+0+0+4+3\lambda=0$
$3\lambda=-4$
$\lambda=-\frac{4}{3}$
Put the value of $\lambda$ in equation $(i),$
$\text{x}(1+\lambda)+\text{y}(2-\lambda)+\text{z}(3+\lambda)+4+3\lambda=0$
$\text{x}\Big(1-\frac{4}{3}\Big)+\text{y}\Big(2+\frac{4}{3}\Big)+\text{z}\Big(3-\frac{4}{3}\Big)+4-\frac{12}{3}=0$
$\text{x}\Big(\frac{3-4}{3}\Big)+\text{y}\Big(\frac{6+4}{3}\Big)+\text{z}\Big(\frac{9-4}{3}\Big)+4-4=0$
$-\frac{\text{x}}{3}+\frac{10\text{y}}{3}+\frac{5\text{z}}{3}=0$
Multiplying by $3,$ we get
$-x + 10y + 5z = 0$
$x - 10y - 5z = 0$
The equation of required plane is,
$x - 10y - 5z = 0$
View full question & answer→Question 1945 Marks
Find the shortest distance between the lines:
$\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}\ \text{and}\ \frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}.$
AnswerEquation of one line is $\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}$
Comparing this equation with $\frac{\text{x}-\text{x}_1}{\text{a}_1}=\frac{\text{y}-\text{y}_1}{\text{b}_1}=\frac{\text{z}-\text{z}_1}{\text{c}_1},$ we have
$x_1 = -1, y_1 = -1, z_1 = -1,$
$a_1 = 7, b_1 = -6, c_1 = 1$
Again equation of another line is $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}$
Comparing this equation with $\frac{\text{x}-\text{x}_2}{\text{a}_2}=\frac{\text{y}-\text{y}_2}{\text{b}_2}=\frac{\text{z}-\text{z}_2}{\text{c}_2},$ we have
$x_2 = 3, y_2 = 5, z_2 = 7,$
$a_2 = 1, b_2 = -2, c_2 = 1$
$\therefore\ \ \begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}$
$=\begin{vmatrix}3+1&5+1&7+1\\7&-6&1\\1&-2&1\end{vmatrix}$
Expanding first row $= 4(-6 + 2) - 6(7 - 1) + 8(-14 + 6) = -16 - 36 - 64 = -116$
And $\sqrt{(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)+(\text{b}_1\text{c}_2-\text{b}_2\text{c}_1)+(\text{c}_1\text{a}_2-\text{c}_2\text{a}_1)}$
$=\sqrt{(-14+6)^2+(-6+2)^2+(1-7)^2}$
$=\sqrt{64+16+36}=\sqrt{116}$
$\therefore$ Length of shortest distance
$=\frac{\begin{vmatrix}\text{x}_2-\text{x}_1&\text{y}_2-\text{y}_1&\text{z}_2-\text{z}_1\\\text{a}_1&\text{b}_1&\text{c}_1\\\text{a}_2&\text{b}_2&\text{c}_2\end{vmatrix}}{\sqrt{(\text{a}_1\text{b}_2-\text{a}_2\text{b}_1)+(\text{b}_1\text{c}_2-\text{b}_2\text{c}_1)+(\text{c}_1\text{a}_2-\text{c}_2\text{a}_1)}}$
$=\frac{-116}{\sqrt{116}}=-\sqrt{116}=\sqrt{116}\ (\text{numerically})$
$=\sqrt{4\times29}=2\sqrt{29}$ View full question & answer→Question 1955 Marks
Find the equations of the planes parallel to the plane x - 2y + 2z - 3 = 0 and which are at a unit distance from the point (1, 1, 1).
AnswerThe equation of the plane parallel to the given plane is
x - 2y + 2z + k = 0 ...(i)
It is given the plane (i) is at a distance of 1 unit from (1, 1, 1)
$\Rightarrow\frac{|1-2+2+\text{k}|}{\sqrt{1^2+(-2)^2+2^2}}=1$
$\Rightarrow\frac{|1+\text{k}|}{3}=1$
⇒ |1 + k| = 3
⇒ 1 + k = 3, 1 + k = -3
Substituting these two values one by one in (i) we get
x - 2y + 2z + 2 = 0 and x - 2y + 2z - 4 = 0, which are the equations of the required planes.
View full question & answer→Question 1965 Marks
Find the equation of the plane that contains the point (1, -1, 2) and is perpendicular to each of the planes 2x + 3y - 2z = 5 and x + 2y - 3z = 8.
AnswerThe equation of any plane passing through (1, -1, 2) is,
a(x - 1) + b(y + 1) + c(z - 2) = 0 ....(i)
It is given that (i) is perpendicular to the plane 2x + 3y - 3z = 5. So,
2a + 3b - 3c = 0 ....(ii)
It is given that (i) is perpendicular to the plane x + 2y - 3z = 8. So,
a + 2b - 3c = 0 ....(iii)
Solving (i), (ii) and (iii) we get,
$\begin{vmatrix}\text{x}-1&\text{y}+1&\text{z}-2\\2&3&-2\\1&2&-3\end{vmatrix}=0$
⇒ -5(x - 1) + 4(y + 1) + 1(z - 2) = 0
⇒ 5x - 4y - z = 7
View full question & answer→Question 1975 Marks
The two adjacent sides of a parallelogram are $2\hat{i} - 4\hat{j} - 5\hat{k} \text{ and } 2\hat{i} + 2\hat{j} + 3\hat{k}.$ Find the two unit vectors parallel its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Answer$\text{let } \text{d}_{1} \& \text{ d}_{2}$ be the two diagonal vectors:
$\therefore \overrightarrow{\text{d}}_{1} = 4\hat{\text{i}} - 2\hat{\text{j}} - 2\hat{\text{k}}, \overrightarrow{\text{d}}_{2} = -6\hat{\text{j}} - 8\hat{\text{k}}$
$\text{or} \overrightarrow{\text{d}}_{2} = 6\hat{\text{j}} + 8\hat{\text{k}}$
Unit vectors parallel to the diagonals are:
$\overrightarrow{\text{d}}_{1} = \frac{2}{\sqrt{6}}\hat{\text{i}} - \frac{1}{\sqrt{6}}\hat{\text{j}} - \frac{1}{\sqrt{6}}\hat{\text{k}}$
$\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} - \frac{4}{5}\hat{\text{k}} \bigg(\overrightarrow{\text{d}}_{2} = \frac{3}{5}\hat{\text{j}} + \frac{4}{5}\hat{\text{k}}\bigg)$
$\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2} = \begin{vmatrix} \hat{\text{i}} & \hat{\text{j}} & \hat{\text{k}} \\ 4 & -2 & -2 \\ 0 & -6 & -8 \end{vmatrix} = 4\hat{\text{i}} + 32\hat{\text{j}} - 24\hat{\text{k}} $
Area of parallelogram $= \frac{1}{2}|\overrightarrow{\text{d}}_{1}\times\overrightarrow{\text{d}}_{2}| = \sqrt{404} \text{ or } 2\sqrt{101} \text{sq. units}$
View full question & answer→Question 1985 Marks
Find the distance of the point (3, 3, 3) from the plane $\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$
AnswerThe given plane is
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})+9=0$
$\vec{\text{r}}\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})=9$
We know that the perpendicular distance of a point P of position vector $\vec{\text{a}}$ from the plane $\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}$ is given by
$\text{p}=\frac{\big|\vec{\text{a}}\cdot\vec{\text{n}}-\text{d}\big|}{|\vec{\text{n}}|}$
Finding the distance from (3, 3, 3) $($which means $3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})$ to the given plane
Here, $\vec{\text{a}}=3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}},\vec{\text{n}}=5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}},\text{d}=9$
So, the required distance p
$=\frac{\big|(3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}})\cdot(5\hat{\text{i}}+2\hat{\text{j}}-7\hat{\text{k}})-9\big|}{\big|3\hat{\text{i}}+3\hat{\text{j}}+3\hat{\text{k}}\big|}$
$=\frac{|-15-6+21-9|}{\sqrt{25+4+49}}$
$=\frac{-9}{\sqrt{78}}$
$=\frac{-9}{\sqrt{78}}\text{ units}$
View full question & answer→Question 1995 Marks
Find the intercepts made on the coordinate axes by the plane 2x + y - 2z = 3 and also find the direction cosines of the normal to the plane.
AnswerHere, given equation of plane is,
2x + y - 2z = 3
Dividing by 3 on both the sides,
$\frac{2\text{x}}{3}+\frac{\text{y}}{3}-\frac{2\text{z}}{3}=\frac{3}{3}$
$\frac{\text{x}}{\frac{3}{2}}+\frac{\text{y}}{3}+\frac{\text{z}}{-\frac{3}{2}}=1\ ...(\text{i})$
We know that, if a, b, c are the intercepts by a plane on the coordinate axes,
new equation of the plane is,
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}+\frac{\text{z}}{\text{c}}=1\ ....(\text{ii})$
Comparing the equation (i) and (ii),
$\text{a}=\frac{3}{2},\text{b}=3,\text{c}=-\frac{3}{2}$
Again, given equation of plane is,
2x + y - 2z = 3
$(\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}})(2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=3$
$\vec{\text{r}}(2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}})=3$
So, vector normal to the plane is given by
$\vec{\text{n}}=2\hat{\text{i}}+\hat{\text{j}}-2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{(2)^2+(1)^2+(-2)^2}$
$=\sqrt{4+1+4}$
$=\sqrt{9}$
$|\vec{\text{n}}|=3$
Direction vector of $\vec{\text{n}}=2,1,-2$
Direction vector $\vec{\text{n}}=\frac{2}{|\vec{\text{n}}|},\frac{1}{|\vec{\text{n}}|},\frac{-2}{|\vec{\text{n}}|}$
$=\frac{2}{3},\frac{1}{3},-\frac{2}{3}$
So,
Intercepts by the plane on coordinaye axes are $=\frac{3}{2},3,-\frac{3}{2}$
Direction cosine of normal to the plane are $=\frac{2}{3},\frac{1}{3},-\frac{2}{3}$
View full question & answer→Question 2005 Marks
Find the equation of a plane which passes through the point (3, 2, 0) and contains the line $\frac{\text{x}-3}{1}=\frac{\text{y}-6}{5}=\frac{\text{z}-4}{4}$
AnswerRequired equation of plane is passing through the point (3, 2, 0)
$\therefore$ a(x - 3) + b(y - 2) + c(z - 0) = 0
⇒ a(x - 3) + b(y - 2) + cz = 0 ....(i)
Required equation of plane also contains the line $\frac{\text{x}-3}{1}=\frac{\text{y}-6}{5}=\frac{\text{z}-4}{4},$ so it passes through the point (3, 2, 0)
⇒ a(3 - 3) + b(6 - 2) + c(4) = 0
⇒ 4b + 4c = 0 ...(ii)
Also plane will be parallel to,
a(1) + b(5) + c(4) = 0
a + 5b + 4c = 0 ....(iii)
Solving (ii) and (iii) by cross multiplication,
$\frac{\text{a}}{16-20}=\frac{\text{b}}{4-0}=\frac{\text{c}}{0-4}=\lambda(\text{say})$
$-\frac{\text{a}}{4}=\frac{\text{b}}{4}=-\frac{\text{c}}{4}=\lambda(\text{say})$
$\Rightarrow\text{a}=-\lambda,\text{ b}=\lambda,\text{ c}=-\lambda$
Put $\text{a}=-\lambda,\text{ b}=\lambda,\text{ c}=-\lambda$ in equation (i) we get
$(-\lambda)(\text{x}-3)+(\lambda)(\text{y}-2)+(-\lambda)\text{z}=0$
$\Rightarrow-\text{x}+3+\text{y}-2-\text{z}=0$
$\Rightarrow\text{x}-\text{y}+\text{z}-1=0$
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