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Maths 2 Marks Questions

VECTOR ALGEBRA · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 215 questions.

Questions · Page 5 of 5

2 Marks Questions

Question 2012 Marks
If $\vec{\text{a}}=2\hat{\text{i}}+\hat{\text{k}},\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}},$ find the magnitude of $\vec{\text{a}}\times\vec{\text{b}}.$
Answer
Given:$\vec{\text{a}}=2\hat{\text{i}}+0\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{b}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&0&1\\1&1&1 \end{vmatrix}$
$=(0-1)\hat{\text{i}}-(2-1)\hat{\text{j}}+(2-0)\hat{\text{k}}$
$=-\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$\Rightarrow\big|\vec{\text{a}}\times\vec{\text{b}}\big|=\sqrt{(-1)^2+(1-)^2+2^2}$
$=\sqrt{6}$
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Question 2022 Marks
Write down a unit vector in XY-plane, making an angle of 30° with the positive direction of x-axis.
Answer
If $\vec{\text{r}}$ is a unit vector in the XY-plane, then $\vec{\text{r}}=\cos\theta\hat{\text{i}}+\text{sin}\theta\hat{\text{j}}.$
Here, $\theta$ is the angle made by the unit vector with the positive direction of the x-axis.
Therefore, for $\theta$ = 30°:
$\vec{\text{r}}=\cos30^{\circ}\hat{\text{i}}+\text{sin}30^{\circ}\hat{\text{j}}=\frac{\sqrt{3}}{2}\hat{\text{i}}+\frac{1}{2}\hat{\text{j}}$
Hence, the required unit vector is $\frac{\sqrt{3}}{2}\hat{\text{i}}+\frac{1}{2}\hat{\text{j}}$
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Question 2032 Marks
If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}\ \text{with}\ \hat{i},\frac{\pi}{4}\ \text{with}\ \hat{j}$ and an acute angle $\theta$ with $\hat{k},$ then find $\theta$ and hence, the components of $\vec{a}.$
Answer
$\text{Let}\ \ \ \hat{a}=\text{x}\hat{i}+\text{y }\hat{j}+\text{z}\hat{k}\ \text{be a unit vector.}\ \ \ \ ......\text{(i)}$ $\Rightarrow\ \ |\hat{a}|=1\ \ \Rightarrow\ \ \sqrt{\text{x}^2+\text{y}^2+\text{z}^2}=1$ $\text{Squaring both sides,}\ \ \ \ \ \text{x}^2+\text{y}^2+\text{z}^2=1\ \ \ .....\text{(ii)}$ $\text{Given:}\ \ \text{Angle between vectors}\ \hat{a}\ \text{and}\ \hat{i}=\hat{i}+0\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{3}.$ $\therefore\ \cos\frac{\pi}{3}=\frac{\hat{a}.\hat{i}}{\big|\hat{a}\big|.\big|\hat{i}\big|}\ $ $\Rightarrow\ \ \frac{1}{2}=\frac{\text{x}(1)+\text{y}(0)+\text{z}(0)}{(1)(1)}$ $\Rightarrow\ \ \ \frac{1}{2}=\text{x}\ \ \ ....\text{(iii)}$Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{j}=0\hat{i}+\hat{j}+0\hat{k}\ \text{is}\ \frac{\pi}{4}.$
$\therefore\ \cos\frac{\pi}{4}=\frac{\hat{a}.{\hat{j}}}{|\hat{a}|.|\hat{j}|}\ \Rightarrow\ \frac{1}{\sqrt{2}}=\frac{\text{x}(0)+\text{y}(1)+\text{z}(0)}{(1)(1)}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\text{y}\ \ ......\text{(iv)}$
Again, given $\text{Angel between vectors}\ \hat{a}\ \text{and}\ \hat{k}=0\hat{i}+0\hat{j}+\hat{k}\ \text{is}\ \theta,\ \text{where}\ \theta\ \text{is acute angle.}$
$\therefore\ \cos\theta=\frac{\hat{a}.{\hat{k}}}{|\hat{a}|.|\hat{k}|}\ $ $\Rightarrow\ \ \cos\theta=\frac{\text{x}(0)+\text{y}(0)+\text{z}(1)}{(1)(1)}$ $\Rightarrow\ \ \cos\theta=\text{z}\ \ \ .......\text{(v)}$ Putting the values of x, y and z in eq. (ii),$\frac{1}{4}+\frac{1}{2}+\cos^2\theta=1$ $\ \Rightarrow\ \ \cos^2\theta=1-\frac{1}{4}-\frac{1}{2}$
$\Rightarrow\ \ \ \ \ \ \ \cos^2\theta=\frac{4-1-2}{4}=\frac{1}{4}$ $\ \Rightarrow\ \ \cos\theta=\pm\frac{1}{2}$ Since $\theta$ is acute angle, therefore cos $\theta$ is positive and hence $\frac{1}{2}=\cos\frac{\pi}{3}\ \Rightarrow\ \ \theta=\frac{\pi}{3}$ From eq. (v), $\ \ \text{z}=\cos\theta=\frac{1}{2}$ Putting values of x, y and z in eq. (i), $\ \ \hat{a}=\frac{1}{2}\hat{i}+\frac{1}{\sqrt{2}}\hat{j}+\frac{1}{2}\hat{k}$$\therefore\ \ \text{Components of}\ \hat{a}\ \text{are coefficients of}\ \hat{i},\hat{j},\hat{k}\ \text{in}\ \hat{a}$
$\Rightarrow\ \ \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2}\ \text{and angle}\ \theta=\frac{\pi}{3}$
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Question 2042 Marks
Evaluate the product $\big(3\vec{a}-5\vec{b})\cdot\big(2\vec{a}+7\vec{b}).$
Answer
$\text{Given:}\ \ \ \ \big(3\vec{a}-5\vec{b}\big).\big(2\vec{a}+7\vec{b}\big)$ $=\big({3\vec{a}\big)}.\big(2\vec{a}\big)+\big(3\vec{a}\big).\big(7\vec{b}\big)-\big(5\vec{b}\big).\big(2\vec{a}\big)-\big(5\vec{b}\big).\big(7\vec{b}\big)$
$=6\vec{a}.\vec{a}+21\vec{a}.\vec{b}-10\vec{b}.\vec{a}-35\vec{b}.\vec{b}$

$=6\big|\vec{a}\big|^2+21\vec{a}.\vec{b}-10\vec{a}.\vec{b}-35\Big|\vec{b}\Big|^2$

$=6\big|\vec{a}\big|^2+11\vec{a}.\vec{b}-35\Big|\vec{b}\Big|^2$
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Question 2052 Marks
If $\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+2\hat{\text{k}},$ find $\big(\vec{\text{a}}-2\vec{\text{b}}\big).\big(\vec{\text{a}}+\vec{\text{b}}\big).$
Answer
We have
$\vec{\text{a}}=\hat{\text{i}}-\hat{\text{j}}$ and $\vec{\text{b}}=-\hat{\text{j}}+2\hat{\text{k}}$
$\vec{\text{a}}-2\vec{\text{b}}=\big(\hat{\text{i}}-\hat{\text{j}}\big)-2\big(-\hat{\text{j}}+2\hat{\text{k}}\big)\\=\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{j}}-4\hat{\text{k}}=\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}$
$\vec{\text{a}}+\vec{\text{b}}=\hat{\text{i}}-\hat{\text{j}}-\hat{\text{j}}+2\hat{\text{k}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$\big(\vec{\text{a}}-2\vec{\text{b}}\big).\big(\vec{\text{a}}+\vec{\text{b}}\big)$
$=\big(\hat{\text{i}}+\hat{\text{j}}-4\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}\big)$
$=1-2-8$
$=-9$
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Question 2062 Marks
Prove that 1, 1, 1 cannot be direction cosines of a straight line.
Answer
Let 1, 1, 1 be the direction cosines of a straight line. Then,
$1^2+1^2+1^2=3\neq1$
Since direction cosines of a line which makes equal angle with the axes must satisfy
$\text{l}^2+\text{m}^2+\text{n}^2=1$
Hence 1, 1, 1 cannot be the direction cosines of a straight line.
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Question 2072 Marks
Find the angle betwwen two vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ if$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
Answer
We have,
$|\vec{\text{a}}|=\sqrt{3},\big|\vec{\text{b}}\big|=2$ and $\vec{\text{a}}.\vec{\text{b}}=\sqrt{6}$
Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}$. then
$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}$
$=\frac{\sqrt{6}}{\sqrt{3}\times2}$
$=\frac{1}{\sqrt{2}}$
$\theta=\cos^{-1}\Big(\frac{1}{\sqrt{2}}\Big)$
$=\frac{\pi}{4}$
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Question 2082 Marks
Show that
$(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=2(\vec{a}\times{\vec{b}})$
Answer
$\text{L.H.S}=(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=\vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b}$ $\ \Big[\because\vec{a}\times\vec{a}=\vec{0},\vec{b}\times\vec{b}=\vec{0},\vec{b}\times\vec{a}=-\vec{a}\times\vec{b}\Big]$
$=\vec{0}+\vec{a}\times\vec{b}+\vec{a}\times\vec{b}-\vec{0}=2\vec{a}\times\vec{b}=\text{R.H.S.}$
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Question 2092 Marks
If $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$, find the unit vector in the direction of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}$.
Answer
Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}},\ \vec{\text{b}}=\hat{\text{j}}+\hat{\text{k}},\ \vec{\text{c}}=\hat{\text{k}}+\hat{\text{i}}$
Then, $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{j}}+\hat{\text{k}}+\hat{\text{k}}+\hat{\text{i}}$
$=2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
$\therefore\ |\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}|=\sqrt{2^2+2^2+2^2}$
$=\sqrt{12}$
$=2\sqrt3$
Therefore, unit vector in the direction of $\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=\frac{2\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)}{2\sqrt3}=\frac{1}{\sqrt3}\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)$
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Question 2102 Marks
Find the angle at which the following vectors are inclined to each of the coordinate axes: $\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$
Answer
Let $\vec{\text{r}}$ be the given vector, and let it make an angle $\alpha,\beta,\gamma$ with OX, OY, OZ respectively. Then, its direction cosines are $\cos\alpha,\cos\beta,\cos\gamma$.So direction ratios of $\vec{\text{r}}=\hat{\text{i}}-\hat{\text{j}}+\hat{\text{k}}$ are proportional to 1, -1, 1. Therefore, Direction cosine of $\vec{\text{r}}$ are $\frac{1}{\sqrt{1^2+(-1)^2+1^2}},\frac{-1}{\sqrt{1^2+(-1)^2+1^2}},\frac{1}{\sqrt{1^2+(-1)^2+1^2}}$ or, $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$ $\therefore\cos\alpha=\frac{1}{\sqrt{3}},\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}$ $\alpha=\cos^{-1}=\Big(\frac{1}{\sqrt{3}}\Big),\beta=\cos^{-1}=\Big(\frac{-1}{\sqrt{3}}\Big),\gamma=\cos^{-1}=\Big(\frac{1}{\sqrt{3}}\Big)$
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Question 2112 Marks
What is the cosine of the angle with the vector $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ makes with y-axis?
Answer
Given $\sqrt2\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$
Therefore, direction cosines are $\frac{\sqrt2}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}},\frac{1}{\sqrt{(\sqrt2)^2+1^2+1^2}}$ or $\frac{1}{\sqrt2},\frac{1}2,\frac{1}2$
So, cosine angle with respect to y-axis is $\frac{1}2$
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Question 2122 Marks
Find the angle between two vectors $\vec{a}\ \text{and}\ \vec{b}$ with magnitudes $\sqrt{3}\ \text{and}\ 2,$ respectively having $\vec{a}\cdot\vec{b}=\sqrt{6}.$
Answer
$\text{Given:}\ \ \big|\vec{a}\big|=\sqrt{3},\Big|\vec{b}\Big|=2\ {\text{and}}\ \vec{a}.\vec{b}=\sqrt{6}$ Let $\theta$ be the angle between the vector $\vec{a}\ \text{and}\ \vec{b}.$ We know that $\text{cos}\ \theta=\frac{\vec{a}.\vec{b}}{\big|\vec{a}\big|.\big|\vec{b}\big|}$$\Rightarrow\ \ \text{cos}\ \theta=\frac{\sqrt{6}}{\sqrt{3}.2}=\frac{\sqrt{3}.\sqrt{2}}{\sqrt{3}.\sqrt{2}.\sqrt{2}}=\frac{1}{\sqrt{2}}$ $\Rightarrow\ \ \text{cos}\ \theta=\text{cos}\frac{\pi}{4}$
$\Rightarrow\ \theta= \frac{\pi}{4}$
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Question 2132 Marks
Show that the direction cosines of a vector equally inclined to the axes $OX, OY$ and $OZ$ are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
Answer
Let a vector be equally inclined to axes OX, OY, and OZ at angle a.
Then, the direction cosines of the vector are cos a, cos a, and cos a.
Now,
$\cos^{2}\alpha+\cos^{2}\alpha+\cos^{2}\alpha=1$
$\Rightarrow3\cos^{2}\alpha=1$
$\Rightarrow\cos\alpha=\frac{1}{\sqrt{3}}$
Hence, the direction cosines of the vector which are equally inclined to the axes are $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}.$
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Question 2142 Marks
Find $\big|\vec{\text{a}}-\vec{\text{b}}\big|$ if
$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=4$
Answer
Given that$|\vec{\text{a}}|=2,\big|\vec{\text{b}}\big|=3$ and $\vec{\text{a}}.\vec{\text{b}}=4\dots(1)$
We know that
$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$
$=2^2+3^2-2(4)$ [using (1)]
$=4+9-8$
$=5$
$\therefore​​​​ \big|\vec{\text{a}}-\vec{\text{b}}\big|=\sqrt{5}$
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Question 2152 Marks
For what of $\lambda$ are the vectors $\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$ perpendicular to each other?
Answer
We have
$\vec{\text{a}}=2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$
Given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are perpendicular.
$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=0$
$\Rightarrow\big(2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}\big)=0$
$\Rightarrow2-2\lambda+3=0$
$\Rightarrow5-2\lambda=0$
$\therefore\lambda=\frac{5}{2}$
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2 Marks Questions - Page 5 - Maths STD 12 Science Questions - Vidyadip