Question 11 Mark
Assertion $(A):$ The lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ are perpendicular, when $\vec{b}_1 \cdot \vec{b}_2=0$.
Reason $(R):$ The angle $\theta$ between the lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by $\cos \theta=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}$.
Answerif lines are perpendicular, then $\theta=\frac{\pi}{2}$
$\therefore \cos \frac{\pi}{2}=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right| \vec{b}_2 \mid}$
$\Rightarrow \cos \frac{\pi}{2}=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right| \vec{b}_2 \mid}$
$\Rightarrow \vec{b}_1 \cdot \vec{b}_2=0$
$\therefore \quad$ Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
View full question & answer→Question 21 Mark
Assertion $(A) :$ Quadrilateral formed by vertices $A(0,0,0), B(3,4,5), C(8,8,8)$ and $D(5,4,3)$ is a rhombus. Reason $(R): A B C D$ is a rhombus if $A B=B C=C D=D A$, $A C \neq B D$.
AnswerGiven, $A(0,0,0), B(3,4,5), C(8,8,8)$ and $D(5,4,3)$
$A B=\sqrt{3^2+4^2+5^2}=\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2} \text { units }$
$B C=\sqrt{(8-3)^2+(8-4)^2+(8-5)^2}=\sqrt{25+16+9}=\sqrt{50}$
$=5 \sqrt{2} \text { units }$
$C D=\sqrt{(5-8)^2+(4-8)^2+(3-8)^2}=\sqrt{9+16+25}=\sqrt{50}=5 \sqrt{2} \text { units }$
$D A=\sqrt{5^2+4^2+3^2}=\sqrt{25+16+9}=\sqrt{50}=5 \sqrt{2} \text { units }$
$A C=\sqrt{8^2+8^2+8^2}=\sqrt{3 \times 8^2}=8 \sqrt{3} \text { units }$
$B D=\sqrt{(5-3)^2+(4-4)^2+(3-5)^2}=\sqrt{4+0+4}=\sqrt{8}=2 \sqrt{2} \text { units }$
$\therefore A B=B C=C D=D A, A C \neq B D$
Hence, both Assertion $(A)$ and Reason $(R)$ are true and Reason $(R)$ is the correct explanation of Assertion $(A).$
View full question & answer→Question 31 Mark
Assertion (A): The pair of lines given by $\vec{r}=\hat{i}-\hat{j}+\lambda(2 \hat{i}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{k}+\mu(\hat{i}+\hat{j}-\hat{k})$ intersect.
Reason $(R) :$ Two lines intersect each other, if they are not parallel and shortest distance $=0$.
AnswerHere, $\vec{a}_1=\hat{i}-\hat{j}, \vec{b}_1=2 \hat{i}+\hat{k}$
$\vec{a}_2=2 \hat{i}-\hat{k}$ and $\vec{b}_2=\hat{i}+\hat{j}-\hat{k}$
$\because \vec{b}_1 \neq k \vec{b}_2$, for any scalar $k$
$\therefore \quad$ Given lines are not parallel.
Now, $\vec{a}_2-\vec{a}_1=(2 \hat{i}-\hat{k})-(\hat{i}-\hat{j})=\hat{i}+\hat{j}-\hat{k}$
and $\vec{b}_1 \times \vec{b}_2=-\hat{i}+3 \hat{j}+2 \hat{k}$
$\Rightarrow\left|\vec{b}_1 \times \vec{b}_2\right|=\sqrt{(-1)^2+(3)^2+(2)^2}=\sqrt{1+9+4}=\sqrt{14}$
$\therefore \text { S.D. }=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|=\left|\frac{(\hat{i}+\hat{j}-\hat{k}) \cdot(-\hat{i}+3 \hat{j}+2 \hat{k})}{\sqrt{14}}\right|=0$
Hence, two lines intersect each other.
Two lines intersect each other, if they are not parallel and shortest distance $=0$.
View full question & answer→Question 41 Mark
Assertion (A) : The points $(1,2,3),(-2,3,4)$ and $(7,0,1)$ are collinear.
Reason (R): If the points $\left(x_1, y_1, z_1\right),\left(x_2, y_2, z_2\right)$ and $\left(x_3, y_3, z_3\right)$ are collinear, then
$\frac{x_2-x_1}{x_3-x_2}=\frac{y_2-y_1}{y_3-y_2}=\frac{z_2-z_1}{z_3-z_2} .$
AnswerWe have, $x_1=1, y_1=2, z_1=3$;
$x_2=-2, y_2=3, z_2=4$ and $x_3=7, y_3=0, z_3=1$
Now, $\frac{x_2-x_1}{x_3-x_2}=\frac{y_2-y_1}{y_3-y_2}=\frac{z_2-z_1}{z_3-z_2}$
$\Rightarrow \frac{-2-1}{7-(-2)}=\frac{3-2}{0-3}=\frac{4-3}{1-4}$
$\Rightarrow \frac{-3}{9}=\frac{1}{-3}=\frac{1}{-3}$
$ \Rightarrow \frac{-1}{3}=\frac{-1}{3}=\frac{-1}{3}$
$\therefore \quad$ Given points are collinear.
Hence, both assertion and reason are true and reason is the correct explanation of assertion.
View full question & answer→Question 51 Mark
Assertion (A) : The lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ are perpendicular, when $\vec{b}_1 \cdot \vec{b}_2=0$.
Reason (R): The angle $\theta$ between the lines $\vec{r}=\vec{a}_1+\lambda \vec{b}_1$ and $\vec{r}=\vec{a}_2+\mu \vec{b}_2$ is given by $\cos \theta=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right|\left|\vec{b}_2\right|}$.
Answer(a) : If lines are perpendicular, then $\theta=\frac{\pi}{2}$
$
\therefore \quad \cos \frac{\pi}{2}=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right| \vec{b}_2 \mid} \Rightarrow \cos \frac{\pi}{2}=\frac{\vec{b}_1 \cdot \vec{b}_2}{\left|\vec{b}_1\right| \vec{b}_2 \mid}
$
$
\Rightarrow \quad \vec{b}_1 \cdot \vec{b}_2=0
$
$\therefore \quad$ Both $A$ and $R$ are true and $R$ is the correct explanation of $A$.
View full question & answer→Question 61 Mark
Assertion (A) : If the cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$, then its vector form is $\vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$.
Reason (R): The vector equation of line passing through the points $A(\vec{a})$ and parallel to vector $\vec{b}$ is $\vec{r}=\vec{a}-\lambda(\vec{a}-\vec{b})$, where $\lambda \in R$ is a parameter.
AnswerIn assertion the given cartesian equation is
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2} .$
$\Rightarrow \vec{a}=5 \hat{i}-4 \hat{j}+6 \hat{k} \text { and } \vec{b}=3 \hat{i}+7 \hat{j}+2 \hat{k} .$
The vector equation of the line is given by $\vec{r}=\vec{a}+\lambda \vec{b}, \lambda \in R$.
$\Rightarrow \vec{r}=5 \hat{i}-4 \hat{j}+6 \hat{k}+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})$
Thus assertion is true and reason is false.
View full question & answer→Question 71 Mark
Directions: In these questions, a statement of Assertion is followed by a statement of Reason is given.Choose the correct answer out of the following choices:
Assertion: If the cartesian equation of a line is $\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2},$ then its vector form is $\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}).$
Reason: The cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by $\frac{\text{x}+3}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+8}{6}$ is $\frac{\text{x}+3}{-2}=\frac{\text{y}-4}{4}=\frac{\text{z}+8}{-5}.$
- Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- Assertion is correct statement but Reason is wrong statement.
- Assertion is wrong statement but Reason is correct statement.
Answer
- Assertion is correct statement but Reason is wrong statement.
Solution:
In assertion the given cartesian equation is
$\frac{\text{x}-5}{3}=\frac{\text{y}+4}{7}=\frac{\text{z}-6}{2},$
$\Rightarrow\vec{\text{a}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}$ and $\vec{\text{b}}=3\hat{\text{i}}+7\hat{\text{j}}+2\hat{\text{k}}$
The vector equation of the line is given by $\vec{\text{r}}=\vec{\text{a}}+\lambda\vec{\text{b}},\lambda\in\text{R}.$
$\Rightarrow\vec{\text{r}}=5\hat{\text{i}}-4\hat{\text{j}}+6\hat{\text{k}}+\lambda(3\hat{\text{i}}+7\hat{\text{j}}+2\text{k})$
Thus Assertion is correct. In reason it is given that the line passes through the point (-2, 4, -5) and is parallel to
Clearly, the direction ratios of line are (3, 5, 6). Now the equation of the line (in cartesian form) is
$\frac{\text{x}-(-2)}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}-(-5)}{6}$
$\Rightarrow\frac{\text{x}+2}{3}=\frac{\text{y}-4}{5}=\frac{\text{z}+5}{6}$ View full question & answer→MCQ 81 Mark
Directions: In the following questions, the Assertions $(A)$ and Reason $(s)(R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: Points $A(4, 0, 4), B(1, 2, 3), C(-2, 4, 2)$ are collinear.
Reason: Three points $A, B, C$ are collinear if $AB + BC = AC$ and $AB, BC < AC.$
- ✓
Both Assertion $\&$ Reason are individually true $\&$ Reason is correct explanation of Assertion.
- B
Both Assertion $\&$ Reason are individually true but Reason is not the, correct $($proper$)$ explanation of Assertion.
- C
Assertion is true but Reason is false.
- D
Assertion is false but Reason is true.
AnswerCorrect option: A. Both Assertion $\&$ Reason are individually true $\&$ Reason is correct explanation of Assertion.
Points $A(4, 0, 4), B(1, 2, 3), C(-2, 4, 2)$ are collinear formula to check whether these three points are collinear or not $AB + BC = AC$ to find $AB$ the equation is
$\sqrt{(\text{x}_2-\text{x}_1)^{2}+(\text{y}_2-\text{y}_1)^2+(\text{z}_1-\text{z}_1)^2}.....(1)$
$x_1 = 4, y_1 = 0$ and $z_1 = 4$
$x_2 = 1, y_2 = 2$ and $z_2 = 3$
by substituting the values in $(1)$ we will get
$AB = 3.7$ similarly for $BC$ and $AC $
$BC = 3.7$
$AC = 7.4$
hence finally its is known that these points are collinear
View full question & answer→Question 91 Mark
Directions: In the following questions, the Assertions (A) and Reason(s) (R) have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion: The points (1, 2, 3), (-2, 3, 4) and (7, 0, 1) are collinear
Reason: If a line makes angles $\frac{\pi}{2}, \frac{3\pi}{4}$ and $\frac{\pi}{4}$ with X, Y, and Z - axes respectively, then its direction cosines are $0,\frac{-1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$
- Assertion and Reason both are correct statements and Reason is the correct explanation of Assertion.
- Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
- Assertion is correct statement but Reason is wrong statement.
- Assertion is wrong statement but Reason is correct statement.
Answer
- Assertion and Reason both are correct statements but Reason is not the correct explanation of Assertion.
Solution:
We have, $\text{x}_1=1,\text{y}_1=2,\text{z}_1=3;$
$\text{x}_2=-2,\text{y}_2=3,\text{z}_2=4$ and $\text{x}_3=7,\text{y}_3=0,\text{z}_3=1$
Now, $\frac{\text{x}_2-\text{x}_1}{\text{x}_3-\text{x}_2}=\frac{\text{y}_2-\text{y}_1}{\text{y}_3-\text{y}_2}=\frac{\text{z}_2-\text{z}_1}{\text{z}_3-\text{z}_2}$
$\Rightarrow\frac{-2-1}{7-(-2)}=\frac{3-2}{0-3}=\frac{4-3}{1-4}$
$\Rightarrow\frac{-3}{9}=\frac{1}{-3}=\frac{1}{-3}\Rightarrow\frac{-1}{3}=\frac{-1}{3}=\frac{-1}{3}$ View full question & answer→Question 101 Mark
Assertion (A) : The point $A(1,0,7)$ is the mirror image of the point $B(1,6,3)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$.
Reason (R) : The line : $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$ bisects the line segment joining $A(1,0,7)$ and $B(1,6,3)$.
Answer(b) : The direction ratios of the line segment joining $A(1,0,7)$ and $B(1,6,3)$ is $(0,6,-4)$.
The direction ratios of the given line is $(1,2,3)$.
As $1 \cdot 0+6 \cdot 2-4 \cdot 3=0$, we have the lines are perpendicular.
Also the midpoint of $A B$ is $(1,3,5)$. Also, the point $(1,3,5)$ lies on the line.
$\therefore \quad$ Point $A$ is the mirror image of point $B$ in the given line. Also, the line bisects $A B$, so statement $I$ and statement II are true.
Statement 'II' holds even if the line is not perpendicular. This situation is possible.
View full question & answer→