1 Marks - Maths STD 12 Science Questions - Vidyadip

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Question 11 Mark

Find the projection of the vector $\vec{a}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ on the vector $\vec{b}=\hat{i}+2 \hat{j}+\hat{k}$.

Answer

$
\begin{array}{l}
\text { Projection of } \vec{a} \text { or } \vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} \\
=\frac{(2 \hat{i}+3 \hat{j}+2 \hat{k}) \cdot(\hat{i}+2 \hat{j}+\hat{k})}{\sqrt{(1)^2+(2)^2+(1)^2}} \\
=\frac{2 \cdot 1+3 \cdot 2+2 \cdot 1}{\sqrt{6}}=\frac{2+6+2}{\sqrt{6}}=\frac{10}{\sqrt{6}}=\frac{5}{3} \sqrt{6}
\end{array}
$

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Question 21 Mark

Find the position vector of the mid-point of the vector joining the points $P (2,3,4)$ and $Q (4,1,-2)$.

Answer

We know that the position vector of joining two points
$P (\vec{a})$ and $Q (\vec{b})$ is $\frac{\vec{a}+\vec{b}}{2}$. Hence according to question,
$
\begin{aligned}
\vec{a} & =2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{b}=4 \hat{i}+\hat{j}-2 \hat{k} \\
& =\frac{(2 \hat{i}+3 \hat{j}+4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2} \\
& =\frac{6 \hat{i}+4 \hat{j}+2 \hat{k}}{2}=3 \hat{i}+2 \hat{j}+\hat{k}
\end{aligned}
$

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Question 31 Mark

The vector directed from A to B and joining the points $A (1,2,2)$ and $B (2,3,1)$ is :

Answer

Vector directed from A towards B
$\begin{aligned}& \overrightarrow{\mathrm{AB}}=(2-1) \hat{i}+(3-2) \hat{j}+(1-2) \hat{k} \\& \overrightarrow{\mathrm{AB}}=\hat{i}+\hat{j}-\hat{k} \end{aligned}$

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Question 41 Mark

The initial point of a vector is $(2,1)$ and the terminal point is $(-5,7)$, find the vectors components of this vector.

Answer

$
\begin{aligned}
\vec{r} & =(-5-2) \hat{i}+(7-1) \hat{j} \\
\vec{r} & =-7 \hat{i}+6 \hat{j}
\end{aligned}
$
Here, $-7 \hat{i}, 6 \hat{j}$ are the vector components of $\vec{r}$ along the coordinate axes.

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Question 51 Mark

In triangle OAC if B is the mid-point of side AC and $\overrightarrow{ OA }=\vec{a}$ and $\overrightarrow{ OB }=\vec{b}$, then what will be $\overrightarrow{ OC }$ ?

Answer

$\begin{aligned} \overrightarrow{ OA }+\overrightarrow{ AB } & =\overrightarrow{ OB } \\ \therefore \quad \overrightarrow{ AB } & =\overrightarrow{ OB }-\overrightarrow{ OA } \\ =\vec{b}-\vec{a} \\ \therefore \quad \overrightarrow{ AB } & =\overrightarrow{ BC } \\ \therefore \quad \overrightarrow{ BC } & =\vec{b}-\vec{a} \\ \overrightarrow{ OC } & =\overrightarrow{ OB }+\overrightarrow{ BC }\end{aligned}$
$=\vec{b}+\vec{b}-\vec{a}=\overrightarrow{2 b}-\vec{a}$

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Question 61 Mark

Define position vector.

Answer

Let $\overrightarrow{ OP }$ be a vector whose initial point O is the fixed point or origin. Then $\overrightarrow{ OP }$ is called the position vector of $P$ which express the position of point $P$ with respect to fixed point $O$. Hence by the position vector the position of a variable point $P$ is determined with respect to fixed point (origin) which is expressed by $\overrightarrow{ OP }$.

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Question 71 Mark

If the vectors $a \hat{i}-a \hat{j}+\hat{k}$ and $a \hat{i}+\hat{j}-2 \hat{k}$ are perpendicular to each other, then find the value of $a$.

Answer

$\because a \hat{i}-a \hat{j}+\hat{k}$ and $a \hat{i}+\hat{j}-2 \hat{k}$ are perpendicular to each other.
$
\begin{array}{rlrl}
\therefore (a \hat{i}-a \hat{j}+\hat{k}) \cdot(a \hat{i}+\hat{j}-2 \hat{k}) =0 \\
\Rightarrow a^2-a-2=0 \Rightarrow(a-2)(a+1) & =0 \\
\Rightarrow =2,-1
\end{array}
$

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Question 81 Mark

If the position vector $a$ of the point $(12, n)$ is such that $|\vec{a}|=13$, then find the value of $n$.

Answer

Position vector of the point $(12, n)=12 \hat{i}+n \hat{j}$
$
\begin{array}{l}
\therefore \quad \vec{a}=12 \hat{i}+n \hat{j} \\
\Rightarrow \quad|\vec{a}|=\sqrt{12^2+n^2} \quad \Rightarrow 13=\sqrt{144+n^2} \\
\Rightarrow \quad 169=144+n^2 \quad \Rightarrow n^2=25 \\
\Rightarrow n= \pm 5
\end{array}
$

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Question 91 Mark

If the coordinates of points P and Q are $(2,-1,1)$ and $(1,-3,-5)$ respectively, then find the vector $\overrightarrow{ PQ }$.

Answer

Let $O$ be the origin.
$
\begin{aligned}
\therefore \quad \text { Position vector } \overrightarrow{OP} =2 \hat{i}-\hat{j}+\hat{k} \\
\text { Position vector } \overrightarrow{OQ} \quad =\hat{i}-3 \hat{j}-5 \hat{k} \\
\therefore \quad \text {PQ} =\overrightarrow{OQ}-\overrightarrow{OP} \\
=\hat{i}-3 \hat{j}-5 \hat{k}-2 \hat{i}+\hat{j}-\hat{k} \\
=-\hat{i}-2 \hat{j}-6 \hat{k} \quad \text { }
\end{aligned}
$

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Question 101 Mark

If $\vec{a}$ is a vector and $m$ is a scalar quantity and $m \vec{a}=\overrightarrow{0}$ then what alternative values of $\vec{a}$ and $m$ are possible?

Answer

Either $m=0$ or $\vec{a}=0$.

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Question 111 Mark

OABC is a tetrahedron. Write the vectors $\overrightarrow{ BC }, \overrightarrow{ CA }, \overrightarrow{ AB }$ in terms of $\overrightarrow{ OA }, \overrightarrow{ OB }$ and $\overrightarrow{ OC }$.

Answer

$\begin{aligned} \overrightarrow{ BC } & =\overrightarrow{ OC }-\overrightarrow{ OB } \\ \overrightarrow{ CA } & =\overrightarrow{ OA }-\overrightarrow{ OC } \\ \overrightarrow{ AB } & =\overrightarrow{ OB }-\overrightarrow{ OA }\end{aligned}$

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Question 121 Mark

Elaborate the unit vector.

Answer

The vector whose magnitude (modulus) is zero, is called zero vector. Its direction is in definite. Those vector whose initial point and terminal point are one and the same, for example $\overrightarrow{ AA }, \overrightarrow{ BB }, \ldots$ etc. always express zero vector. It is shown by $\overrightarrow{0}$ or black zero $(\overrightarrow{0})$.

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Question 131 Mark

In the adjacent figure, find the value of $\overrightarrow{ OB }$.

Answer

By triangle law of vector addition
$
\begin{aligned}
\overrightarrow{OB} & =\overrightarrow{OA}+\overrightarrow{AB} \\
& =\vec{a}+\vec{b}
\end{aligned}
$

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Question 141 Mark

In $\triangle A B C$, write the sum of three vectors represented by $\overrightarrow{ AB }, \overrightarrow{ BC }, \overrightarrow{ CA }$ respectively.

Answer

$
\begin{aligned}
\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA} & =(\overrightarrow{AB}+\overrightarrow{BC})+\overrightarrow{CA} \\
& =\overrightarrow{AC}+\overrightarrow{CA} \\
& =\overrightarrow{AC}-\overrightarrow{AC}=\overrightarrow{0}
\end{aligned}
$

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Question 151 Mark

Find the unit vector in the direction of $2 \hat{i}+2 \hat{j}+\hat{k}$.

Answer

$
\begin{aligned}
\text { Unit vector } & =\frac{\text { Vector }}{\mid \text { Vector } \mid} \\
& =\frac{2 \hat{i}+2 \hat{j}+\hat{k}}{\sqrt{(2)^2+(2)^2+(1)^2}} \\
& =\frac{1}{3}(2 \hat{i}+2 \hat{j}+\hat{k})
\end{aligned}
$

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1 Marks - Maths STD 12 Science Questions - Vidyadip