M.C.Q (1 Marks)

Question 1511 Mark

find the coordinate of the tip of the position vector which is equivalent to $\overrightarrow{\text{AB}}$ where the coordinates of A and B are (-1, 3) and (-2, 1) respectively:

(+1, +2)
(+1, -2)
(-1, +2)
(-1, -2)

Answer

(-1, -2)

View full question & answer→

Question 1521 Mark

If $\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1.$then $\vec{\text{a}}=$

$\vec{0}$
$\hat{\text{i}}$
$\hat{\text{j}}$
$\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$

Answer

$\hat{\text{i}}$

Solution:
Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$
$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$
and $\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$
and $\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$
Given,
$\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1$
$\Rightarrow\text{a}_1=\text{a}_1+\text{a}_2=\text{a}_1+\text{a}_2+\text{a}_3=1$
$\Rightarrow\text{a}_1=1,\text{a}_2=0,\text{a}_3=0$
So, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$

View full question & answer→

Question 1531 Mark

If the position vectors of P and Q are $\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}$ and $5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}$ then the cosine of the angle getween $\overrightarrow{\text{PQ}}$ and y-axis is:

$\frac{5}{\sqrt{162}}$
$\frac{4}{\sqrt{162}}$
$-\frac{5}{\sqrt{162}}$
$\frac{11}{\sqrt{162}}$

Answer

$-\frac{5}{\sqrt{162}}$

Solution:
$\overrightarrow{\text{PQ}}=\overrightarrow{\text{OQ}}-\overrightarrow{\text{OP}}=5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}-\big(\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}\big)=4\hat{\text{i}}-5\hat{\text{j}}+11\hat{\text{k}}$
The unit vector along y-axis is $\hat{\text{j}}.$
Let $\theta$ be the required angle.
$\cos\theta=\frac{\overrightarrow{\text{PQ}}.\hat{\text{j}}}{\big|\overrightarrow{\text{PQ}}\big|\big|\hat{\text{j}}\big|}=\frac{-5}{\sqrt{16+25+121}\sqrt{1}}=\frac{-5}{\sqrt{162}}$

View full question & answer→

Question 1541 Mark

If $\hat{\text{i}},\hat{\text{j}},\hat{\text{k}}$ are unit vectors, then

$\hat{\text{i}}.\hat{\text{j}}=1$
$\hat{\text{i}}.\hat{\text{i}}=1$
$\hat{\text{i}}\times\hat{\text{j}}=1$
$\hat{\text{i}}\times\big(\hat{\text{j}}\times\hat{\text{k}}\big)=1$

Answer

$\hat{\text{i}}.\hat{\text{i}}=1$

Solution:
Let us check each option one by one.

We know

$\hat{\text{i}}.\hat{\text{j}}=0$
$\neq1$

We know

$\hat{\text{i}}.\hat{\text{i}}=|\hat{\text{i}}|^2$
$=1^2$
$=1$

$\hat{\text{i}}\times\hat{\text{j}}=\hat{\text{k}}$

$\neq1$

$\hat{\text{i}}\times\big(\hat{\text{i}}\times\hat{\text{k}}\big)=\hat{\text{i}}\times\hat{\text{i}}$

$=0$
$\neq1$

View full question & answer→

Question 1551 Mark

The direction cosines l, m and n of two lines are connected by the relations l + m + n = 0, l m = 0, then the angles between them is:

$\frac{\pi}{3}$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
0

Answer

$\frac{\pi}{3}$

View full question & answer→

MCQ 1561 Mark

The equation of normal to the curve $3x^2 - y^2 = 8$ which is parallel to the line $x + 3y = 8$ is$:$

A
$3x - y = 8$
B
$3x + y + 8 = 0$
✓
$\text{x + 3y} \underline{+} 8 = 0$
D
$x + 3y = 0$

Answer

Correct option: C.

$\text{x + 3y} \underline{+} 8 = 0$

View full question & answer→

Question 1571 Mark

Which of the following holds true for a vector quantity:

It has only magnitude
It has only direction
A vector has both direction and magnitude
A vector can never be negative

Answer

A vector has both direction and magnitude

Solution:
A quantity which has both magnitude and direction is called a vector quantity. The quantity which has only magnitude but no direction is called a scalar quantity.

View full question & answer→

Question 1581 Mark

If y = x:

0.32
0.032
5.68
5.968

Answer

0.32

View full question & answer→

Question 1591 Mark

If $\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ and $\vec{\text{d}}$ are the position vector of points A, B, C, D such that no three of them are collinear and $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$, then ABCD is a,

Rhombus.
Rectangle.
Square.
Parallelogram.

Answer

Parallelogram

Solution:
Given:
$\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\vec{\text{c}}-\vec{\text{d}}=\vec{\text{b}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}$
And $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\vec{\text{c}}-\vec{\text{b}}=\vec{\text{d}}-\vec{\text{a}}$
$\Rightarrow\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}$
Also, since $\vec{\text{a}}+\vec{\text{c}}=\vec{\text{b}}+\vec{\text{d}}$
$\Rightarrow\frac{1}2\big(\vec{\text{a}}+\vec{\text{c}}\big)=\frac{1}2\big(\vec{\text{b}}+\vec{\text{d}}\big)$
So, position vector of mid-point of BD = position vector of mid-point of AC.
Hence diagonals bisect each other.
The given ABCD is a parallelogram.

View full question & answer→

Question 1601 Mark

If $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}}$ are three non-zero vectors, no two of which are collinear and the vector $\vec{\text{a}}+\vec{\text{b}}$ is collinear with $\vec{\text{c}}$, $\vec{\text{b}}+\vec{\text{c}}$ is collinear with $\vec{\text{a}}$, then $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=$

$\vec{\text{a}}$
$\vec{\text{b}}$
$\vec{\text{c}}$
None of these

Answer

None of these

Solution:
$\vec{\text{a}}+\vec{\text{b}}$ is collinear with $\vec{\text{c}}$
$\therefore\ \vec{\text{a}}+\vec{\text{b}}=\text{x}\vec{\text{c}}\ \dots(1)$
where x is scalar and $\text{x}\neq0$.
$\vec{\text{b}}+\vec{\text{c}}$ is collinear with $\vec{\text{a}}$
$\vec{\text{b}}+\vec{\text{c}}=\text{y}\vec{\text{a}}\ \dots(2)$
y is scalar and $\text{y}\neq0$
Substracting (2) from (1) we get,
$\vec{\text{a}}-\vec{\text{c}}=\text{x}\vec{\text{c}}-\text{y}\vec{\text{a}}$
$\vec{\text{a}}(1+\text{y})=(1+\text{x})\vec{\text{c}}$
As given $\vec{\text{a}},\ \vec{\text{c}}$ are not collinear,
$\therefore\ 1+\text{y}=0$ and $1+\text{x}=0$
$\text{y}=-1$ and $\text{x}=-1$
Putting the value of x in equation (1)
$\vec{\text{a}}+\vec{\text{b}}=-\vec{\text{c}}$
$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$

View full question & answer→

Question 1611 Mark

If $\theta$ is the angle between any two vectors $\vec{\text{a}}$ and $\vec{\text{b}},$ then $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ when $\theta$ is equal to:

$0$
$\frac{\pi}{4}$
$\frac{\pi}{2}$
$\pi$

Answer

$\frac{\pi}{4}$

Solution:
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$
We know
$\big|\vec{\text{a}}\times\vec{\text{b}}\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$
$\Rightarrow\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big||\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\big|=|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|$
Given: $\big|\vec{\text{a}}.\vec{\text{b}}\big|=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$
$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big||\cos\theta|=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\sin\theta$
$\Rightarrow|\cos\theta|=\sin\theta$
$\Rightarrow\theta=\frac{\pi}{4}$

View full question & answer→

Question 1621 Mark

If points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear, then a is equal to,

Answer

Solution:
Given: Three points $\text{A}(60\hat{\text{i}}+3\hat{\text{j}}),\ \text{B}(40\hat{\text{i}}-8\hat{\text{j}})$ and $\text{C}(\text{a}\hat{\text{i}}+52\hat{\text{j}})$ are collinear. Then,
$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
We have,
$\overrightarrow{\text{AB}}=\big(40\hat{\text{i}}-8\hat{\text{j}}\big)-\big(60\hat{\text{i}}+3\hat{\text{j}}\big)$
$=-20\hat{\text{i}}-11\hat{\text{j}}$
$\overrightarrow{\text{BC}}=\big(\text{a}\hat{\text{i}}-52\hat{\text{j}}\big)-\big(40\hat{\text{i}}-8\hat{\text{j}}\big)$
$=(\text{a}-40)\hat{\text{i}}-44\hat{\text{j}}$
Therefore,
$\overrightarrow{\text{AB}}=\lambda\overrightarrow{\text{BC}}$
$\Rightarrow-20\hat{\text{i}}-11\hat{\text{j}}=\lambda(\text{a}-40)\hat{\text{i}}-\lambda44\hat{\text{j}}$
$\Rightarrow\lambda(\text{a}-40)=-20,\ -44\lambda=-11\Rightarrow\lambda=\frac{1}4$
$\Rightarrow\text{a}-40=-80$
$\Rightarrow\text{a}=-40$

View full question & answer→

Question 1631 Mark

The distance of the point (-3, 4, 5) from the origin:

50
$5\sqrt{2}$
6
None of these

Answer

$5\sqrt{2}$

View full question & answer→

Question 1641 Mark

If $\mid\text{a}\mid=4$ and $-3\underline{<}\lambda\underline{<}2$ then the range of $\mid\lambda\text{a}\mid$ is:

[0, 8]
[-12, 8]
[0, 12]
[8, 12]

Answer

[0, 12]

View full question & answer→

Question 1651 Mark

If the vectors $3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$ are perpendicular, then $\lambda$ is equal to:

$-14$
$7$
$14$
$\frac{1}{7}$

Answer

$14$

Solution:
It is given that vectors $3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$ are perpendicular.
So, their dot product is zero.
$\big(3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}\big)=0$
$\Rightarrow6-\lambda+8=0$
$\Rightarrow14-\lambda=0$
$\therefore\lambda=14$

View full question & answer→

Question 1661 Mark

Choose the correct answer from the given four options.
The vectors $\lambda\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\ \hat{\text{i}}+\lambda\hat{\text{j}}-\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$ are coplanar if:

$\lambda=-2$
$\lambda=0$
$\lambda=1$
$\lambda=-1$

Answer

$\lambda=-2$

Solution:
Let $\vec{\text{a}}=\lambda\hat{\text{i}}+\hat{\text{j}}+2\hat{\text{k}},\ \vec{\text{b}}=\hat{\text{i}}+\lambda\hat{\text{j}}-\hat{\text{k}}$ and $\vec{\text{c}}=2\hat{\text{i}}-\hat{\text{j}}+\lambda\hat{\text{k}}$
For $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ to be coplanar,
$\begin{vmatrix}\lambda&1&2 \\1&\lambda&-1\\2&-1&\lambda \end{vmatrix}=0$
$\Rightarrow\lambda(\lambda^2-1)-1(\lambda+2)+2(-1-2\lambda)=0$
$\Rightarrow\lambda^3-\lambda-\lambda-2-2-4\lambda=0$
$\Rightarrow\lambda^3-6\lambda-4=0$
$\Rightarrow(\lambda+2)(\lambda^2-2\lambda-2)=0$
$\Rightarrow\lambda=-2$ or $\lambda=\frac{2\pm\sqrt{12}}{2}$
$\Rightarrow\lambda=-2$ or $\lambda=1\pm\sqrt{3}$

View full question & answer→

MCQ 1671 Mark

The equation of tangent to the curve $y(1 + x^2) = 2 - x, w$ here it crosses $x-$axis is$:$

✓
$x + 5y = 2$
B
$x - 5y = 2$
C
$5x - y = 2$
D
$5x + y = 2$

Answer

Correct option: A.

$x + 5y = 2$

View full question & answer→

Question 1681 Mark

The vector $(\cos\text{a}\cos\beta)\hat{\text{i}}+(\cos\text{a}\sin\beta)\hat{\text{j}}+(\sin\text{a})\hat{\text{k}}$is a:

Null vector
Unit vector
Constant vector
None of these

Answer

Unit vector

Solution:
Let $\vec{\text{a}}=(\cos\text{a}\cos\beta)\hat{\text{i}}+(\cos\text{a}\sin\beta)\hat{\text{j}}+(\sin\text{a})\hat{\text{k}}$
$|\vec{\text{a}}|=\sqrt{\cos^2\text{a}\cos^2\beta+\cos^2\text{a}\sin^2\beta+\sin^2\text{a}}$
$=\sqrt{\cos^2\text{a}(\cos^2\beta+\sin^2\beta)+\sin^2\text{a}}$
$=\sqrt{\cos^2\text{a}(1)+\sin^2\text{a}}$
$=\sqrt{\cos^2\text{a}+\sin^2\text{a}}$
$=\sqrt{1}$
$=1$
So, $\vec{\text{a}}$ is a unit vector.

View full question & answer→

Question 1691 Mark

In triangle ABC (Fig 10.18), which of the following is not true:

$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec{0}$
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec{0}$
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec{0}$
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec{0}$

Answer

On applying the triangle law of addition in the given triangle, we have:
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}\ \ \ \ \ \ \ \ \ ....(1)$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=-\overrightarrow{\text{CA}}$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}+\overrightarrow{\text{CA}}=\vec{0}\ \ \ \ ....(2)$
$\therefore$ The equation given in alternative A is true.
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{AC}}$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AC}}=\vec{0}$
$\therefore$ The equation given in alternative B is true.
From equation (2), we have:
$\overrightarrow{\text{AB}}-\overrightarrow{\text{CB}}+\overrightarrow{\text{CA}}=\vec{0}$
$\therefore$ The equation given in alternative D is true.
Now, consider the equation given in alternative C:
$\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{CA}}=\vec{0}$
$\Rightarrow\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}=\overrightarrow{\text{CA}}\ \ \ \ ....(3)$
From equations (1) and (3), we have:
$\overrightarrow{\text{AC}}=\overrightarrow{\text{CA}}$
$\Rightarrow\overrightarrow{\text{AC}}=-\overrightarrow{\text{AC}}$
$\Rightarrow\overrightarrow{\text{AC}}+\overrightarrow{\text{AC}}=\vec{0}$
$\Rightarrow2\overrightarrow{\text{AC}}=\vec{0}$
$\Rightarrow\overrightarrow{\text{AC}}=\vec{0},$ which is not true.
Hence, the equation given in alternative C is incorrect.
The correct answer is C.

View full question & answer→

Question 1701 Mark

If $\vec{\text{a}}$ is a non-zero of magnitude 'a' and $\lambda$ is a non-zero scalar, then $\lambda\vec{\text{a}}$ is a unit vector if:

$\lambda=1$
$\lambda=-1$
$\text{a}=|\lambda|$
$\text{a}=\frac{1}{|\lambda|}$

Answer

$\text{a}=\frac{1}{|\lambda|}$

Solution:
Given that
$|\vec{\text{a}}|=\text{a};$
Now,
$|\lambda\vec{\text{a}}|=1$
$\Rightarrow|\lambda||\vec{\text{a}}|=1$
$\Rightarrow|\lambda|\text{a}=1$
$\Rightarrow\text{a}=\frac{1}{|\lambda|}$

View full question & answer→

Question 1711 Mark

ABCD is a parallelogram with AC and BD as diagonals. Then, $\overrightarrow{\text{AC}}-\overrightarrow{\text{BD}}=$

$4\overrightarrow{\text{AB}}$
$3\overrightarrow{\text{AB}}$
$2\overrightarrow{\text{AB}}$
$\overrightarrow{\text{AB}}$

Answer

$2\overrightarrow{\text{AB}}$

Solution:
Given: ABCD, a parallelogram with diagonals AC and BD. Then,
$\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}$
$\overrightarrow{\text{AD}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BD}}$
$\Rightarrow\ \overrightarrow{\text{BD}}=\overrightarrow{\text{AD}}-\overrightarrow{\text{AB}}$
$\therefore\overrightarrow{\text{AC}}-\overrightarrow{\text{BD}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}-\overrightarrow{\text{AD}}+\overrightarrow{\text{AB}}=2\overrightarrow{\text{AB}}$ $\Big[\because\ \overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}\Big]$

View full question & answer→

Question 1721 Mark

Choose the correct answer from the given four options.
The number of vectors of unit length perpendicular to the vectors $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{j}}+\hat{2\text{k}}$ and $\vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}}$ is:

One.
Two.
Three.
Infinite.

Answer

Two.

Solution:
The number of vectors of unit length perpendicular to the vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ is $\vec{\text{c}}$ (say) i.e., $\vec{\text{c}}=\pm(\vec{\text{a}}\times\vec{\text{b}})$
So, there will be two vectors of unit length perpendicular to the vectors $\vec{\text{a}}$ and $\vec{\text{b}}.$

View full question & answer→

MATHS M.C.Q (1 Marks)