In the article, we will solve Miscellaneous Exercise from Chapter 10, “Vector Algebra” in the NCERT. This exercise covers the basics of vectors like scalar and vector components of vectors, section formulas, Multiplication of a Vector by a scalar, etc.
Question 1: Write down a unit vector in XY-plan, making an angle of 30 degree with the positive direction of x-axis
Answer
Let us take \vec{r} as a unit vector in the XY-plan, then \vec{r}=cos\theta\hat{i}+sin\theta {\hat{j}}
Also,\theta is the angle made by the unit vector with the positive direction of x-axis.
Therefore, for \theta=30\degree{}:
\vec{r}=cos30\degree{}\hat{i}+sin30\degree{} {\hat{j}}=\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}
Hence, the required unit vector is\frac{\sqrt{3}}{2}\hat{i}+\frac{1}{2}\hat{j}
Question 2: Find the scalar components and magnitude of the vector joining the points
P(x_1,y_1,z_1)\, and \, Q(x_2,y_2,z_2)
Answer
The vector joining the points P(x_1,y_1,z_1)\, and \, Q(x_2,y_2,z_2) can be obtained by,
\vec{{PQ}} = Position vector of Q-Position vector of P
=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{z}
\left|\vec{PQ}\right|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
Hence, the scalar components and magnitude of the vector joining the points are:
\{(x_2-x_1),(y_2-y_1),(z_2-z_1)\}\,and\,\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
Question 3: A girl walks 4 km towards west, then she walks 3 km in a direction 30\degree east of north and stops. Determine the girl's displacement from her initial point of departure.
Answer
Let O and B be the initial and final positions of the girl respectively.
Now, we have
\vec{OA}=-4\hat{i}\\ \vec{AB}=\hat{i}\left|\vec{AB}\right|cos60\degree+\hat{j}\left|\vec{AB}\right|sin60\degree\\=\hat{i}3\times\frac{1}{2}+\hat{j}3\times\frac{\sqrt{3}}{2}
=\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
Also, by the triangle law of vector addition we have
\vec{OB}=\vec{OA}+\vec{AB}
=(-4\hat{i})+(\frac{3}{2}\hat{i}+\frac{3\sqrt{3}}{2}\hat{j})
=(-4+\frac{3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
=(\frac{-8+3}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
=(\frac{-5}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
Hence, the girl's displacement from her initial point of departure is
=(\frac{-5}{2})\hat{i}+\frac{3\sqrt{3}}{2}\hat{j}
Question 4: if \vec{a}=\vec{b}+\vec{c}, then is it true that \left|\vec{a}\right|=\left|\vec{b}\right|+\left|\vec{c}\right|\,? Justify your answer
Answer:
In\, \triangle ABC,let\,\vec{CB}=\vec{a},\vec{CA}=\vec{b}\,and \,\vec{AB}=\vec{c}
By the triangle law of vector addition, we have \vec{a}=\vec{b}+\vec{c}
Also, we know \left|\vec{a}\right|,\left|\vec{b}\right|,\left|\vec{c}\right|\, represent the sides of \triangle ABC.
The sum of the lengths of any two sides of a triangle is greater than the third side.
\left|\vec{a}\right|< \left|\vec{b}\right|+\left|\vec{c}\right|\,
Hence, it is not true that \left|\vec{a}\right|=\left|\vec{b}\right|+\left|\vec{c}\right|\,.
Question 5: Find the value of x for which x(\hat{i}+\hat{j}+\hat{k}) is a unit vector.
Answer:
We know x(\hat{i}+\hat{j}+\hat{k}) is a unit vector if \left|x(\hat{i}+\hat{j}+\hat{k})\right|=1
Now,
\left|x(\hat{i}+\hat{j}+\hat{k})\right|=1
=\sqrt{x^2+x^2+x^2}=1\\=\sqrt{3x^2}=1\\=\sqrt{3}x=1\\=x= \pm {}_{}\frac{1}{\sqrt{3}}
Hence the required value of x is ±\frac{1}{\sqrt{3}}
Question 6: Find a vector of magnitude 5 units, and parallel to the resultant of the vectors
\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\,and \,\vec{b}=\hat{i}-2\hat{j}+\hat{k}
Answer:
We have,
\vec{a}=2\hat{i}+3\hat{j}-\hat{k}\,and \,\vec{b}=\hat{i}-2\hat{j}+\hat{k}
Let \vec{c} be the resultant of \vec{a}\,and \, \vec{b}
Now,
\vec{c}=\vec{a}+\vec{b}=(2+1)\hat{i}+(3-2)\hat{j}+(-1+1)\hat{k}=3\hat{i}+\hat{j}
\left|\vec{c}\right|=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}
\hat{c}=\frac{\vec{c}}{\left|\vec{c}\right|}=\frac{3\hat{i}+\hat{j}}{\sqrt{10}}
Hence, the vector of magnitude 5 units and parallel to the resultant of the given vectors \vec{a}\,and\,\vec{b} is
\pm{} 5\cdot\hat{c}=\pm 5\cdot \frac{1}{\sqrt{10}}(3\hat{i}+\hat{j})=\pm \frac{3\sqrt{10}\hat{i}}{2}\pm \frac{\sqrt{10}}{2}\hat{j}
Question 7: \vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}\,and\,\vec{c}=\hat{i}-2\hat{j}+\hat{k},\,find\, a\, unit\, vector\,parallel\,to\,thevector\,2\vec{a}-\vec{b}+3\vec{c}
Answer:
We have,
\vec{a}=\hat{i}+\hat{j}+\hat{k},\vec{b}=2\hat{i}-\hat{j}+3\hat{k}\,and\,\vec{c}=\hat{i}-2\hat{j}+\hat{k}
2\vec{a}-\vec{b}+3\vec{c}=2(\hat{i}+\hat{j}+\hat{k})-(2\hat{i}-\hat{j}+3\hat{k})\,+3(\hat{i}-2\hat{j}+\hat{k})
=2\hat{i}+2\hat{j}+2\hat{k}-2\hat{i}+\hat{j}-3\hat{k}\,+3\hat{i}-6\hat{j}+3\hat{k}
=3\hat{i}-3\hat{j}+2\hat{k}
\left|2\vec{a}-\vec{b}+3\vec{c}\right|=\sqrt{3^2+(-3)^2+2^2}=\sqrt{9+9+4}=\sqrt{22}
Therefore, the unit vector along 2\vec{a}-\vec{b}+3\vec{c}\,\,is
\frac{2\vec{a}-\vec{b}+3\vec{c}}{\left |2\vec{a}-\vec{b}+3\vec{c}\right|}=\frac{3\hat{i}-3\hat{j}+2\hat{k}}{\sqrt{22}}
=\frac{3}{\sqrt{22}}\hat{i}-\frac{3}{\sqrt{22}}\hat{j}+\frac{2}{\sqrt{22}}\hat{k}
Question 8: Show that the points A (1, -2, -8), B (5, 0, -2) and C (11, 3, 7) are collinear, and find the ratio in which B divides AC.
Answer:
The given points are A (1, -2, -8), B (5, 0, -2), and C (11, 3, 7).
\vec{AB}=(5-1)\hat{i}+(0+2)\hat{j}+(-2+8)\hat{k}=4\hat{i}+2\hat{j}+6\hat{k}
\vec{BC}=(11-5)\hat{i}+(3-0)\hat{j}+(7+2)\hat{k}=6\hat{i}+3\hat{j}+9\hat{k}
\vec{AC}=(11-1)\hat{i}+(3+2)\hat{j}+(7+8)\hat{k}=10\hat{i}+5\hat{j}+15\hat{k}
\left|\vec{AB}\right|=\sqrt{4^2+2^2+6^2}=\sqrt{16+4+36}=\sqrt{56}=2\sqrt{14}
\left|\vec{BC}\right|=\sqrt{6^2+3^2+9^2}=\sqrt{36+9+81}=\sqrt{126}=3\sqrt{14}
\left|\vec{AC}\right|=\sqrt{{10}^2+5^2+{15}^2}=\sqrt{100+25+225}=\sqrt{350}=5\sqrt{14}
\left|\vec{AC}\right|=\left|\vec{AB}\right|+\left|\vec{BC}\right|
Hence, the given points A, B, and C are collinear.
Now, let point B divide AC in the ratio \lambda:1 then we have:
\vec{OB}=\frac{\lambda \vec{OC}+\vec{OA}}{(\lambda+1)}
=> 5\hat{i}-2\hat{k}=\frac{\lambda (11\hat{i}+3\hat{j}+7\hat{k})+(\hat{i}-2\hat{j}-8\hat{k})}{(\lambda+1)}
=> (\lambda+1)(5\hat{i}-2\hat{k})=11\lambda\hat{i}+3\lambda\hat{j}+7\lambda\hat{k}+\hat{i}-2\hat{j}-8\hat{k}
=>5(\lambda+1)\hat{i}-2(\lambda+1)\hat{k}=(11\lambda+1)\hat{i}+(3\lambda-2)\hat{j}+(7\lambda-8)\hat{k}
On equating the corresponding components, we get:
5(\lambda+1)=11\lambda+1
=> 5\lambda+5=11\lambda+1\\=>6\lambda=4
=>\lambda=\frac{4}{6}=\frac{2}{3}
Hence, point B divides AC in the ratio 2:3
Question 9 Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are (2\vec{a}+\vec{b})and(\vec{a}-3\vec{b}) externally in the ratio 1: 2. Also, show that P is the mid-point of the line segment RQ.
Answer:
Given \vec{OP}=2\vec{a}+\vec{b},\vec{OQ}=\vec{a}-3\vec{b}.
Also, point R divides a line segment joining two points P and Q externally in. the ratio 1: 2. By the section formula, we get:
\vec{OR}=\frac{2(2\vec{a}+\vec{b})-(\vec{a}-3\vec{b})}{2-1}=\frac{4\vec{a}+2\vec{b}-\vec{a}+3\vec{b}}{1}=3\vec{a}+5\vec{b}
Hence, the positive vector of point R is 3\vec{a}+5\vec{b}.
Positive vector of the mid-point of RQ = \frac{\vec{OQ}+\vec{OR}}{2}
=\frac{(\vec{a}-3\vec{b})+(3\vec{a}+5\vec{b})}{2}=2\vec{a}+\vec{b}=\vec{OP}
Hence p is the mid-point of the line Segment RQ
Question 10: The two adjacent sides of a parallelogram are 2\hat{i}-4\hat{j}+5\hat{k}\,\,\,and\,\,\,\hat{i}-2\hat{j}-3\hat{k} Find the unit vector parallel to its diagonal. Also, find its area.
Answer:
Two adjacent sides of a parallelogram are: \vec{a}=2\hat{i}-4\hat{j}+5\hat{k}\,\,\,and\,\,\,\vec{b}=\hat{i}-2\hat{j}-3\hat{k}
The diagonal of parallelogram is given by \vec{a}+\vec{b}
\vec{a}+\vec{b} = (2+1)\hat{i}+(-4-2)\hat{j}+(5-3)\hat{k}=3\hat{i}-6\hat{j}+2\hat{k}
Thus, the unit vector parallel to the diagonal is
\frac{\vec{a}+\vec{b}}{\left|\vec{a}+\vec{b}\right|}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{3^2+(-6)^2+2^2}}
\frac{3\hat{i}-6\hat{j}+2\hat{k}}{\sqrt{9+36+4}}=\frac{3\hat{i}-6\hat{j}+2\hat{k}}{7}=\frac{3}{7}\hat{i}-\frac{6}{7}\hat{j}+\frac{2}{7}\hat{k}
Area of parallelogram ABCD = \left|\vec{a}\times\vec{b}\right|
\vec{a}\times\vec{b}=\begin{vmatrix} \hat{i} && \hat{j}&& \hat{k}&\\ \\ 2&&-4&&5&\\ \\ 1&&-2&&-3&\\ \end{vmatrix}
=\vec{i}(12+10)-\vec{j}(-6-5)+k(-4+4)=22\hat{i}+11\hat{j}=11(\hat{i}+\hat{j})
\left|\vec{a}*\vec{b}\right|=11\sqrt{2^2+1^2}=11\sqrt{5}
Hence, the area of the parallelogram is 11\sqrt{5} square units.
Question 11: 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:
Consider a vector is 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^2a+cos^2a+cos^2a=1
=>3cos^2a=1
= cosa=\frac{1}{\sqrt{3}}
Therefore, 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}}
Question 12:Let\,\,\vec{a}=\hat{i}+4\hat{j}+2\hat{k},\vec{b}=3\hat{i}-2\hat{j}+7\hat{k}\,\, and\,\,\,\vec{c}=2\hat{i}-\hat{j}+4\hat{k}. Find a vector \vec{d} which is perpendicular to both \vec{a}\,\,and\,\,\vec{b},and\,\,\vec{c}.\vec{d}=15
Answer:
Let\,\,\vec{d}=d_1\hat{i}+d_2\hat{j}+d_3\hat{k}
We know \vec{d} is perpendicular to both \vec{a}\,\,and\,\,\vec{b},we\,\,have:\\ \vec{d}.\vec{a}=0\\ = d_1+4d_2+2d_3=0 \,\,\,\,---(i)
Also,
\vec{d}.\vec{b}=0\\ => 3d_1-2d_2+7d_3=0\,\,\,---(ii)
And
\vec{c}.\vec{d}=15\\ => 2d_1-d_2+4d_3=15\,\,\,---(iii)
Solving (i),(ii),and (iii),we get:
d_1=\frac{160}{3},d_2=-\frac{5}{3}\,\,and\,\,d_3=-\frac{70}{3}
\vec{d}=\frac{160}{3}\hat{i}-\frac{5}{3}\hat{j}-\frac{70}{3}\hat{k}\\ =>\frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})
Therefore, the required vector is \frac{1}{3}(160\hat{i}-5\hat{j}-70\hat{k})
Question 13: The scalar product of the vector \hat{i}+\hat{j}+\hat{k} with a unit vector along the sum of vectors 2\hat{i}+4\hat{j}-5\hat{k}\,\,and\,\,\lambda\hat{i}+2\hat{j}+3\hat{k} is equal to one. Find the value of λ.
Answer:
(2\hat{i}+4\hat{j}-5\hat{k})+(\,\,\lambda\hat{i}+2\hat{j}+3\hat{k})
=(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}
Therefore, unit vector along (2\hat{i}+4\hat{j}-5\hat{k})+(\,\,\lambda\hat{i}+2\hat{j}+3\hat{k}) \,\,is\,\,given \,\,as:
\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{4+4\lambda+\lambda^2+36+4}}
=\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{\lambda^2+4\lambda+44}}
Scalar product of (\hat{i}+\hat{j}+\hat{k}) with its unit vector is 1.
=>(\hat{i}+\hat{j}+\hat{k}).\frac{(2+\lambda)\hat{i}+6\hat{j}-2\hat{k}}{\sqrt{\lambda^2+4\lambda+44}}=1
=>\frac{(2+\lambda)+6-2}{\sqrt{\lambda^2+4\lambda+44}}=1
=>\sqrt{\lambda^2+4\lambda+44}=\lambda+6
=>\lambda^2+4\lambda+44=(\lambda+6)^2
=>\lambda^2+4\lambda+44=\lambda^2+12\lambda+36\\ =>8\lambda=8\\=>\lambda=1
Therefore, the value of \lambda is 1.
Question 14: if \vec{a},\vec{b},\vec{c} mutually perpendicular vectors of equal magnitudes are, show that the vector \vec{a}+\vec{b}+\vec{c} equally inclined to \vec{a},\vec{b},\,\,and\,\,\vec{c}.
Answer:
Given that \vec{a},\vec{b},\,\,and\,\,\vec{c}are mutually perpendicular vectors, Hence we have
\vec{a}.\vec{b}=\vec{b}.\vec{c}=\vec{c}.\vec{a}=0
Also,
\left|\vec{a}\right|=\left|\vec{b}\right|=\left|\vec{c}\right|
The vector \vec{a}+\vec{b}+\vec{c} be inclined to \vec{a},\vec{b}\,\,and\,\,\vec{c} at angle \theta_1,\theta_2,\,\,and\,\,\theta_3\,\,respectively.
Now we have:
cos\theta_1=\frac{(\vec{a}+\vec{b}+\vec{c}).\vec{a}}{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{a}\right|}
=\frac{\vec{a}.\vec{a}+\vec{b}.\vec{a}+\vec{c}.\vec{a}}{{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{a}\right|}}
=\frac{\left|\vec{a}\right|^2}{{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{a}\right|}}\,\,\,\,\,\,\,\, [\vec{b}.\vec{a}=\vec{c}.\vec{a}=0]
=\frac{\left|\vec{a}\right|}{{\left|\vec{a}+\vec{b}+\vec{c}\right|}}
cos\theta_2=\frac{(\vec{a}+\vec{b}+\vec{c}).\vec{b}}{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{b}\right|}
=\frac{\vec{a}.\vec{b}+\vec{b}.\vec{b}+\vec{c}.\vec{a}}{{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{b}\right|}}
=\frac{\left|\vec{b}\right|^2}{{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{b}\right|}}\,\,\,\,\,\,\,\, [\vec{a}.\vec{b}=\vec{c}.\vec{b}=0]
=\frac{\left|\vec{b}\right|}{{\left|\vec{a}+\vec{b}+\vec{c}\right|}}
cos\theta_3=\frac{(\vec{a}+\vec{b}+\vec{c}).\vec{c}}{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{c}\right|}
=\frac{\vec{a}.\vec{c}+\vec{b}.\vec{c}+\vec{c}.\vec{c}}{{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{c}\right|}}
=\frac{\left|\vec{c}\right|^2}{{\left|\vec{a}+\vec{b}+\vec{c}\right|\left|\vec{c}\right|}}\,\,\,\,\,\,\,\, [\vec{a}.\vec{c}=\vec{b}.\vec{c}=0]
=\frac{\left|\vec{c}\right|}{{\left|\vec{a}+\vec{b}+\vec{c}\right|}}
Now, as \left|\vec{a}\right|=\left|\vec{b}\right|=\left|\vec{c}\right|,cos\theta_1=cos\theta_2=cos\theta_3.\\\theta_1=\theta_2=\theta_3
Hence, the vector (\vec{a}+\vec{b}+\vec{c}) is equally inclined to \vec{a},\vec{b}\,\,and\,\,\vec{c}.
Question 15: Prove that (\vec{a}+\vec{b}).(\vec{a}+\vec{b})=\left|\vec{a}\right|^2+\left|\vec{b}\right|^2,if only if \vec{a},\vec{b} are perpendicular, given \vec{a}\neq\vec{0},\vec{b}\neq\vec{0}.
Answer:
(\vec{a}+\vec{b}).(\vec{a}+\vec{b})=\left|\vec{a}\right|^2+\left|\vec{b}\right|^2
By \ distributivty \ of \ scalar \ products \ over \ addition\ = \vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}=\left|\vec{a}\right|^2+\left|\vec{b}\right|^2\,\,\,\,\,\,\,
=>\left|\vec{a}\right|^2+2\vec{a}.\vec{b}+\left|\vec{b}\right|^2=\left|\vec{a}\right|^2+\left|\vec{b}\right|^2\,\,\,\,\,\,\,\,\,\,[\vec{a}.\vec{b}=\vec{b}.\vec{a}(Scalar product is commutative)]
=> 2\vec{a}.\vec{b}=0 \\ =>\vec{a}.\vec{b}=0
So\,\, \vec{a}\,\, and\,\,\vec{b}\,\,\, are\,\,\, perpendicular\,\,\,\,\,\,[\vec{a}\ne\vec{0},\vec{b}\ne\vec{0}(Given)]
Question 16: if \theta is the angle between two vectors \vec{a} and \vec{b},then \vec{a}.\vec{b}\geq0\,only\,\,when
(A) \,0<\theta<\frac{\pi}{2}
(B) \,0 \le \theta \le\frac{\pi}{2}
(C) \,0<\theta<\pi
(D) \,0\le\theta \le\pi
Answer:
Let θ be the angle between two vectors\vec{a} \ and\ \vec{b}
Now \vec{a}.\vec{b} = |\vec{a}||\vec{b}| cos\theta
\vec{a}.\vec{b} \geq 0
= cos\theta\geq0
= 0\leq\theta\leq\frac{\pi}{2}
Correct answer is (B) \,0 \le \theta \le\frac{\pi}{2}
Question 17: Let \vec{a} and \vec{b} be two-unit vectors a and \theta is the angle between them.Then \vec{a}+\vec{b} is a unit vector if
(A)\,\,\theta=\frac{\pi}{4}\\ (B)\,\,\theta=\frac{\pi}{3}\\(C)\,\,\theta=\frac{\pi}{2}\\(D)\,\,\theta=\frac{2\pi}{3}
Answer:
We have two unit vectors \vec{a} \ and \ \vec{b} with angle θ between them.
Then, |\vec{a}|\ =\ |\vec{b}| \ = 1
Also if |\vec{a}+\vec{b}| = 1\ then\ \vec{a}+\vec{b}\ is\ a \ unit\ vector.
|\vec{a}+\vec{b}| = 1\\=(\vec{a}+\vec{b})^2 = 1\\=(\vec{a}+\vec{b})\cdot(\vec{a}+\vec{b}) =1 \\ =\vec{a}\cdot\vec{a} + \vec{a}\cdot\vec{b} +\vec{b}\cdot\vec{a}+\vec{b}\cdot\vec{b} = 1 \\ = |\vec{a}|^2 + 2\vec{a}\cdot\vec{b}+|\vec{a}|^2 = 1 \\ =1^2 + 2|\vec{a}||\vec{b}|cos\theta +1^2 = 1 \\ = 1 + 2 \cdot1\cdot1 cos\theta +1=1 \\ cos\theta= -\frac{1}{2} \\ \theta = \frac{2\pi}{3}
Hence, Correct answer is (D)\,\,\theta=\frac{2\pi}{3}
Question 18: The value of \hat{i}.(\hat{j}*\hat{k})+\hat{j}.(\hat{i}*\hat{k})+\hat{k}.(\hat{i}*\hat{j})\,\,is
(A)\,\,0\\(B)\,\,-1\\ (C)\,\,1\\(D)\,\,3
Answer:
\vec{i}\cdot(\vec{j} \times \vec{k}) +\vec{j}\cdot(\vec{i} \times \vec{k})+\vec{k}\cdot(\vec{i} \times \vec{j}) \\ = \vec{i}\cdot\vec{i} + \vec{j}\cdot(-\vec{j})+ \vec{k}\cdot\vec{k} \\ = 1 - 1 + 1 \\= 1
Therefore, correct answer is (C)\,\,1
Question 19: If \theta is the angle between any two vectors \vec{a}\,\,\,and \,\,\,\vec{b}, then \left|\vec{a}.\vec{b} \right|=\left|\vec{a} *\vec{b} \right| when \theta is equal to
(A)\,0 \,\,\, (B)\,\frac{\pi}{4}\,\,(C)\,\frac{\pi}{2}\,\,\,(D)\,\pi
Answer:
Let θ be the angle between \vec{a} \ and \ \vec{b}
\vec{a} \ and \ \vec{b} are non-zero vectors, so |\vec{a} |\ and \ |\vec{b}| \ are \ positive.
|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}| \\= |\vec{a}| |\vec{b}|cos\theta = |\vec{a}||\vec{b}|sin\theta \\ = cos\theta = sin\theta \\= tan\theta = 1 \\ \theta = \frac {\pi}{4}
Hence, Correct answer is (B)\,\frac{\pi}{4}
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