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Maths 1 Marks Question

Vector Algebra · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 15 questions.

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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
Projection of $\vec{a}$ or $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
$=\frac{(2 \hat{i}+3 \hat{j}+2 \hat{k}) \cdot(\hat{i}+2 \hat{j}+\hat{k})}{\sqrt{(1)^2+(2)^2+(1)^2}}$
$=\frac{2 \cdot 1+3 \cdot 2+2 \cdot 1}{\sqrt{6}}$
$=\frac{2+6+2}{\sqrt{6}}$
$=\frac{10}{\sqrt{6}}$
$=\frac{5}{3} \sqrt{6}$
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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 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 41 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 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

Image
$\begin{aligned} \overrightarrow{ OA }+\overrightarrow{ AB } & =\overrightarrow{ OB } \\ \therefore \quad \overrightarrow{ AB } & =\overrightarrow{ OB }-\overrightarrow{ OA } \\ =\vec{b}-\vec{a} \\ \therefore \quad \overrightarrow{ AB } & =\overrightarrow{ BC } \\ \therefore \quad \overrightarrow{ BC } & =\vec{b}-\vec{a} \\ \overrightarrow{ OC } & =\overrightarrow{ OB }+\overrightarrow{ BC }\end{aligned}$
$=\vec{b}+\vec{b}-\vec{a}=\overrightarrow{2 b}-\vec{a}$
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Question 61 Mark
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 71 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 $(12, n)=12 \hat{i}+n \hat{j}$
$\therefore \vec{a}=12 \hat{i}+n \hat{j}$
$\Rightarrow |\vec{a}|=\sqrt{12^2+n^2} \Rightarrow 13=\sqrt{144+n^2}$
$\Rightarrow 169=144+n^2 \Rightarrow n^2=25$
$\Rightarrow n= \pm 5$
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Question 81 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 91 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 101 Mark
OABC is a tetrahedron. Write the vectors $\overrightarrow{ BC }, \overrightarrow{ CA }, \overrightarrow{ AB }$ in terms of $\overrightarrow{ OA }, \overrightarrow{ OB }$ and $\overrightarrow{ OC }$.
Image
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 111 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 121 Mark
In the adjacent figure, find the value of $\overrightarrow{ OB }$.
Image
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 131 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 141 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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Question 151 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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