CBSE Class 12 Maths Chapter 10 Vector Algebra Important Questions
Question 1. Find the angle between the vectors \((\hat{\mathrm{i}}-\hat{\mathrm{j}})\) and \((\hat{\mathrm{j}}-\hat{\mathrm{k}}) \text {. }\)
Or,
Write the projection of the vector \({\vec{r}}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\) on
- x-axis and
- y-axis
Solution:
Let θ be the angle between \(\vec{a}=\hat{i}-\hat{j}\) and \(\vec{b}=\hat{j}-\hat{k}\)
Now, \(\vec{a} \cdot \vec{b}=(\hat{i}-\hat{j}) \cdot(\hat{j}-\hat{k})=-1,|\vec{a}|=\sqrt{2},|\vec{b}|=\sqrt{2}\)
∴ \(\cos \theta=-\frac{1}{2}=\cos \frac{2 \pi}{3} \Rightarrow \theta=\frac{2 \pi}{3}\)
Or,
Given, \({\vec{r}}=3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}\)
Projection of \(\vec{a}\) on \(\vec{b}\) is given as \(\vec{a}\).\(\vec{b}\)
Read and Learn More CBSE Class 12 Maths Important Question and Answers
⇒ Projection of \(\vec{r}\) on x-axis = \(\frac{\vec{r} \cdot \hat{i}}{|\hat{i}|}=\frac{(3 \hat{i}-4 \hat{j}+12 \hat{k}) \hat{i}}{1}=3\)
and projection of \(\vec{r}\) on y-axis = \(\frac{{\mathrm{r}} \cdot \hat{\mathrm{j}}}{|\hat{\mathrm{j}}|}=\frac{(3 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+12 \hat{\mathrm{k}}) \cdot \hat{\mathrm{j}}}{1}=-4\)
Question 2. If \(\vec{a}=\alpha \hat{i}+3 \hat{j}-6 \hat{k} \text { and } \vec{b}=2 \hat{i}-\hat{j}-\beta \hat{k} \text {. }\), find the value of α and β so that \(\vec{a}\) and \(\vec{b}\) may be collinear.
Solution:
⇒ \(\vec{a}=\alpha \hat{i}+3 \hat{j}-6 \hat{k}, \vec{b}=2 \hat{i}-\hat{j}-\beta \hat{k} \) (given)
Since \(\vec{a}\) and \(\vec{b}\) be collinear, we must have: \(\alpha \hat{i}+3 \hat{j}-6 \hat{k}=t(2 \hat{i}-\hat{j}-\beta \hat{k})\)
On comparing both sides
α = 2t, t = -3, -βt = -6
⇒ β = -2, α = 2 x (-3) = -6
Hence α= -6, β = -2
Question 3. Find the magnitude of vector \(\vec{a}\) given by \(\vec{a}\) = \((\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})\)
Solution:
⇒ \({\vec{a}}=(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}) \times(-\hat{\mathrm{i}}+3 \hat{\mathrm{k}})\)
=\(\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
1 & 3 & -2 \\
-1 & 0 & 3
\end{array}\right|\)
= \(9 \hat{\mathrm{i}}-\hat{\mathrm{j}}+3 \hat{\mathrm{k}}\)
= \(|{\vec{a}}|=\sqrt{(9)^2+(-1)^2+(3)^2}=\sqrt{91} \text { units }\)
Question 4. If \(|\vec{a}|=4 \text { and }-3 \leq \lambda \leq 2\), then \(|\lambda \vec{a}|\) lies in:
- [0,12]
- [2,3]
- [8,12]
- [-12,8]
Solution: 1. [0,12]
⇒ \(|\vec{a}|=4\) (given)
Now, \(|\lambda \vec{a}|=|\lambda| \vec{a}|\Rightarrow| \lambda \vec{a}|=4| \lambda\) (because \(|\vec{a}|=4\))
Also, \(-3 \leq \lambda \leq 2\) (given)
⇒ \(0 \leq|\lambda| \leq 3 \Rightarrow 0 \leq 4|\lambda| \leq 12 \Rightarrow 0 \leq|\lambda \vec{a}| \leq 12\)
∴ \(|\lambda \vec{a}| \in[0,12]\)
Question 5. The area of a triangle formed by verticles O, A and B where, \(\overrightarrow{\mathrm{OA}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) and \(\overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k}\), is
- 3√5 sq units
- 5√5 sq units
- 6√5 sq units
- 4 sq units
Solution: 1. 3√5 sq units
⇒ \(\overrightarrow{O B}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{O B}=-3 \hat{i}-2 \hat{j}+\hat{k}\) (given)
Area of \(\triangle \mathrm{OAB}=\frac{1}{2}|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|\)…..(1)
Now, \(\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}\)
= \(\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ -3 & -2 & 1\end{array}\right|=8 \hat{i}-10 \hat{j}+4 \hat{k}\)
⇒ \(|\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}|=\sqrt{64+100+16}=\sqrt{180}\)
So, area of \(\triangle \mathrm{OAB}=\frac{1}{2} \sqrt{180}=3 \sqrt{5}\) sq. units [from (1)]
Question 6. Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude such that the angle between them is 60 and their scalar product is \(\frac{9}{2}\)
Solution:
Given, \(|\vec{a}|=|\vec{b}|\)
Also, angle between \(\vec{a}\) and \(\vec{b}\) is \(60^{\circ}\) and \(\vec{a} \cdot \vec{b}=\frac{9}{2}\)
⇒ \(|\vec{a}||\vec{b}| \cos 60^{\circ}=\frac{9}{2}\)
or \(|\vec{a}|^2 \times \frac{1}{2}=\frac{9}{2}\)
(because \(|\vec{a}|=|\vec{b}|\))
⇒ \(|\vec{a}|^2=9 \text { or }|\vec{a}|=3\)
Hence, magnitudes of \(\vec{a}\) and \(\vec{b}\) are equal to 3 .
Question 7. Write the projection of the vector \((\vec{b}+\vec{c})\) on the vector \(\vec{a}\), where \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \({\vec{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \text {. }\)
Solution:
Given, \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \({\vec{c}}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}+4 \hat{\mathrm{k}}\)
⇒ \(\vec{\mathrm{b}}+\vec{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+2 \hat{\mathrm{k}}\)
∴ Projection of \(\vec{\mathrm{b}}+\vec{\mathrm{c}}\) on \(\vec{\mathrm{a}}\) = \(\vec{\mathrm{b}}+\vec{\mathrm{c}}\) \(\hat{\mathrm{a}}\)
= \((3 \hat{i}+\hat{j}+2 \hat{k}) \cdot \frac{(2 \hat{i}-2 \hat{j}+\hat{k})}{3}=\frac{6}{3}=2 \text { units }\)
Question 8. If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three mutually perpendicular unit vectors, find the value of |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|.
Solution:
Since \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are unit vectors, we have:
⇒ |\(\vec{a}\)| = |\(\vec{b}\)|=|\(\vec{c}\)|= 1….(1)
Since \(\vec{a}\).\(\vec{b}\).\(\vec{a}\) are mutually perpendicular vectors, we have:
⇒ \(\vec{a}\).\(\vec{b}\) = \(\vec{b}\).\(\vec{c}\) = \(\vec{c}\).\(\vec{a}\) = 0…..(2)
Now, |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|2 = (\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{a}\)).(\(\vec{a}\) + 2\(\vec{b}\) +3\(\vec{a}\)) (|\(\vec{a}\)|’=\(\vec{a}\).\(\vec{a}\))
= \(\vec{a}\).\(\vec{a}\) + 4(\(\vec{b}\).\(\vec{b}\))+9(\(\vec{c}\).\(\vec{c}\))+4\(\vec{a}\).\(\vec{b}\)+6\(\vec{a}\).\(\vec{c}\)+12\(\vec{a}\).\(\vec{c}\) (\(\vec{a}\).\(\vec{a}\) = \(\vec{b}\),\(\vec{a}\))
= |\(\vec{a}\)|2 + 4|\(\vec{b}\)|2 + 9|\(\vec{c}\)|2 + 0 [using(2)]
= 1 + 4 + 9 = 14 [using (1)]
⇒ |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|2= 14
⇒ |\(\vec{a}\) + 2\(\vec{b}\) + 3\(\vec{c}\)|= V14
Question 9. If the side AB and BC of a parallelogram ABCD are represented as vectors \(\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\overrightarrow{\mathrm{BC}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\). then find the unit vector along diagonal AC.
Solution:
⇒ \(\overrightarrow{\mathrm{AB}}=2 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}-5 \hat{k}, \overrightarrow{\mathrm{BC}}=\hat{i}+2 \hat{j}+3 \hat{k}\) (given)
⇒ \(\overrightarrow{\mathrm{AC}}=\overrightarrow{\mathrm{AB}}+\mathrm{BC}=(2 \hat{i}+4 \hat{j}-5 \hat{k})+(\hat{i}+2 \hat{j}+3 \hat{k})=3 \hat{i}+6 \hat{j}-2 \hat{k}\)
⇒ \(|\overrightarrow{\mathrm{AC}}|=\sqrt{9+36+4}=\sqrt{49}=7\)
∴ Required unit vector along diagonal \(\mathrm{AC}=\frac{\overrightarrow{\mathrm{AC}}}{|\overrightarrow{\mathrm{AC}}|}\)
= \(\frac{3 \hat{i}+6 \hat{j}-2 \hat{k}}{7}=\frac{1}{7}(3 \hat{i}+6 \hat{j}-2 \hat{k})\)
Question 10. Find a vector 7 equally inclined to the three axes and whose magnitude is 3√3 units.
Or,
Find the angle between unit vectors \(\vec{a}\) and \(\vec{b}\) so that v3\(\vec{a}\)– \(\vec{b}\) is also a unit vector.
Solution:
Let α be the angle made by a vector with coordinate axes.
Then, \({\vec{r}}=3 \sqrt{3}(\ell \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}})\) (because \(|{\vec{r}}|=3 \sqrt{3}\))
where l = cos α, m = cos α, n = cos α
∴ \(l^2+m^2+n^2=1 \Rightarrow 3 \cdot \cos ^2 \alpha=1 \Rightarrow \cos \alpha= \pm \frac{1}{\sqrt{3}}\)
∴ \({\vec{r}}=3 \sqrt{3}\left( \pm \frac{\hat{\mathrm{i}}}{\sqrt{3}} \pm \frac{\hat{\mathrm{j}}}{\sqrt{3}} \pm \frac{\hat{\mathrm{k}}}{\sqrt{3}}\right) \text { or } \overrightarrow{\mathrm{r}}= \pm 3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}\)
Or,
Given, \(\vec{a}\) and \(\vec{b}\) are unit vectors i.e. |\(\vec{a}\)| — 1 — |\(\vec{b}\)|
and \(|\sqrt{3} \vec{a}-\vec{b}|=1 \Rightarrow|\sqrt{3} \vec{a}-\vec{b}|^2=1\)
⇒ \((\sqrt{3} \vec{a}-\vec{b}) \cdot(\sqrt{3} \vec{a}-\vec{b})=1\)
or \(3|\vec{a}|^2-\sqrt{3} \vec{a} \cdot \vec{b}-\sqrt{3} \cdot \vec{a}+|\vec{b}|^2=1\)
⇒ \(3-2 \sqrt{3} \vec{a} \cdot \vec{b}+1=1\)
or \(\vec{a} \cdot \vec{b}=\frac{\sqrt{3}}{2} \Rightarrow|\vec{a}||\vec{b}| \cos \theta=\frac{\sqrt{3}}{2}\)
⇒ \(\cos \theta=\frac{\sqrt{3}}{2} \Rightarrow \theta=\frac{\pi}{6}\)
Question 11. Show that the points \(\mathrm{A}(-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\), \(\mathrm{B}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})\) and \(\mathrm{C}(7 \hat{\mathrm{i}}-\hat{\mathrm{k}})\) are collinear.
Or,
Find \(|\vec{a} \times \vec{b}| \text {, if } \vec{a}=2 \hat{i}+\hat{j}+3 \hat{k} \text { and } \vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k} \text {. }\)
Solution:
⇒ \(\vec{\mathrm{a}}=-2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{k}, \overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \overrightarrow{\mathrm{c}}=7 \hat{\mathrm{i}}-\hat{\mathrm{k}}\) (given)
⇒ \(\overrightarrow{\mathrm{AB}}=\vec{\mathrm{b}}-\vec{\mathrm{a}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\),
⇒ \(\overrightarrow{\mathrm{AC}}=\vec{\mathrm{c}}-\vec{\mathrm{a}}=9 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}=33 \hat{\mathrm{i}}-\hat{\mathrm{j}}-2 \hat{\mathrm{k}}\)
⇒ \(\overrightarrow{\mathrm{AC}}=3 \cdot \overrightarrow{\mathrm{AB}}\)
⇒ \(\overrightarrow{\mathrm{AC}} \| \overrightarrow{\mathrm{AB}}\) but A is common in both \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AB}}\)
⇒ \(\overrightarrow{\mathrm{AC}}\) and \(\overrightarrow{\mathrm{AB}}\) are collinear
⇒ \(\mathrm{A} \cdot \mathrm{B}\) and \(\mathrm{C}\) are collinear
Given, \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}\)
⇒ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right|-\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3) \Rightarrow \vec{a} \times \vec{b}=-17 \hat{i}+13 \hat{j}+7 \hat{k}\)
or \(|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+(7)^2}=\sqrt{289+169+49}=\sqrt{507}=13 \sqrt{3}\)
Question 12. If θ is the angle between two vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-2 \hat{j}+\hat{k}\). find \(\sin \theta\).
Solution:
Let \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}\)
∴ \(\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{\hat{i}-2 \hat{j}+3 \hat{k} \cdot 3 \hat{i}-2 \hat{j}+\hat{k}}{\sqrt{(1)^2+(-2)^2+(3)^2} \times \sqrt{(3)^2+(-2)^2+(1)^2}}=\frac{3+4+3}{\sqrt{14} \cdot \sqrt{14}}=\frac{10}{14}=\frac{5}{7}\)
Hence, \(\cos \theta=\frac{5}{7} \Rightarrow \sin \theta=\sqrt{1-\cos ^2 \theta}=\sqrt{1-\frac{25}{49}}\)
Hence; \(\sin \theta=\sqrt{\frac{24}{49}}=\frac{2}{7} \sqrt{6}\)
Question 13. The two adjacent sides of a parallelogram are represented by vectors \(2 \hat{i}-4 \hat{j}+5 \hat{k}\) and \(\hat{i}-2 \hat{j}-3 \hat{k}\). Find the unit vector parallel to one of its diagonals. Also, find the area of the parallelogram.
Or,
If \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k} \cdot \vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}\) are such that the vector (\(\vec{a}\) + X\(\vec{b}\)). is perpendicular to vector c . then find the value of X.
Solution:
Adjacent sides of a parallelogram are given as \(\vec{a}=2 \hat{i}-4 \hat{j}+5 \hat{k} \text { and } \vec{b}=\hat{i}-2 \hat{j}-3 \hat{k} \text {. }\)
Then, the diagonal of a parallelogram is given by \(\vec{a}\) + \(\vec{b}\)
⇒ \(\vec{a}+\vec{b}=(2 \hat{i}-4 \hat{j}+5 \hat{k})+(\hat{i}-2 \hat{j}-3 \hat{k})=3 \hat{i}-6 \hat{j}+2 \hat{k}\)
Thus, the unit vector parallel to the diagonal is \(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{3 \hat{i}-6 \hat{j}+2 \hat{k}}{\sqrt{3^2+(-6)^2+2^2}}=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}\)
Also, Area of parallelogram = \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|\)
⇒ \(\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}=\left|\begin{array}{ccc}
\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\
2 & -4 & 5 \\
1 & -2 & -3
\end{array}\right|=\hat{\mathrm{i}}(12+10)-\hat{\mathrm{j}}(-6-5)+\hat{\mathrm{k}}(-4+4)\)
⇒ \(\vec{\mathrm{a}} \times \vec{\mathrm{b}}=22 \hat{\mathrm{i}}+11 \hat{\mathrm{j}}\)
∴ \(|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|=\sqrt{22^2+11^2}=11 \sqrt{5}\)
Hence, the area of the parallelogram is 11√5 square units.
Given, \(\vec{\mathrm{a}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}, \vec{\mathrm{b}}=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) and \(\vec{\mathrm{c}}=3 \hat{\mathrm{i}}+\hat{\mathrm{j}}\).
Given, \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\).
⇒ \((\vec{a}+\lambda \vec{b}) \cdot \vec{c}=0 \Rightarrow \vec{a} \cdot \vec{c}+\lambda(\vec{b} \cdot \vec{c})=0 \Rightarrow \lambda=-\frac{\vec{a} \cdot \vec{c}}{b \cdot c}\)
⇒ \(\lambda=-\frac{(2 \hat{i}+2 \hat{j}+3 \hat{k}) \cdot(3 \hat{i}+\hat{j})}{(-\hat{i}+2 \hat{j}+\hat{k}) \cdot(3 \hat{i}+\hat{j})}\)
⇒ \(\lambda=-\frac{6+2+0}{-3+2+0}=8\)
Question 14. If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three vectors such that \(\vec{a}\).\(\vec{b}\) = \(\vec{a}\).\(\vec{c}\) and \(\vec{a}\)x\(\vec{b}\) = \(\vec{a}\)x\(\vec{c}\), \(\vec{a}\)*\(\vec{0}\), then show that \(\vec{b}\) = \(\vec{c}\).
Or,
If |\(\vec{a}\)| = 3. |\(\vec{a}\)| = 5. |\(\vec{c}\)| = 4 and \(\vec{a}\) + \(\vec{b}\) + \(\vec{c}\) = 0. then find the value of (\(\vec{a}\).\(\vec{b}\) + \(\vec{b}\).\(\vec{c}\) + \(\vec{c}\).\(\vec{a}\)).
Solution:
Given; \(\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}\) and \(\vec{a} \neq \vec{0}\)
⇒ \(\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0\) and \(\vec{a} \neq \vec{0}\)
⇒ \(\vec{a} \cdot(\vec{b}-\vec{c})=0\) and \(\vec{a} \neq \vec{0}\)
or \(\vec{b}-\vec{c}=\vec{0}\) or \(\vec{a} \perp(\vec{b}-\vec{c})\)
⇒ \(\vec{b}=\vec{c}\) or \(\vec{a} \perp(\vec{b}-\vec{c})\)
Again ; \(\vec{\mathrm{a}} \times \vec{\mathrm{b}}=\vec{\mathrm{a}} \times \vec{\mathrm{c}}\) and \(\vec{\mathrm{a}} \neq \vec{0}\)
⇒ \((\vec{a} \times \vec{b})-(\vec{a} \times \vec{c})=\vec{0}\) and \(\vec{a} \neq \vec{0}\)
or \(\vec{\mathrm{a}} \times(\vec{\mathrm{b}}-\vec{\mathrm{c}})=\vec{0}\) and \(\vec{\mathrm{a}} \neq \vec{0}\)
⇒ \((\vec{b}-\vec{c})=\vec{0}\) or \(\vec{a} \|(\vec{b}-\vec{c})\)
⇒ \(\vec{b}=\vec{c}\) or \(\vec{a} \|(\vec{b}-\vec{c})\)
From (1) and (2); we get \(\vec{b}\) = \(\vec{c}\)
(\(\vec{a}\) cannot be both and to (\(\vec{b}\)–\(\vec{c}\) simultaneously)
Given that \(|\vec{a}|=3,|\vec{b}|=5,|\vec{c}|=4 and \vec{a}+\vec{b}+\vec{c}=\vec{0}\)
⇒ \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})=0\) (because \(|\vec{a}|^2=\vec{a} \cdot \vec{a}\))
⇒ \(|\vec{a}|^2+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}+\vec{b} \cdot \vec{a}+|\vec{b}|^2+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}+|\vec{c}|^2=0\)
⇒ \(9+25+16+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0\)
(because \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\))
⇒ \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-25\)
Question 15. The scalar product of the vector \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) with a unit vector along the sum of the vectors \(\vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{\mathrm{c}}=\lambda \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}}\) is equal to 1. Find the value of λ and hence find the unit vector along \(\vec{b}\) + \(\vec{c}\).
Solution:
⇒ \(\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{c}=\lambda \hat{i}+2 \hat{j}+3 \hat{k}\)
⇒ \({\vec{b}}+\vec{\mathrm{c}}=\hat{\mathrm{i}}(2+\lambda)+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}=\vec{\mathrm{d}}\)(let)
Unit vector along \(\vec{d}= \pm \frac{\hat{i}(2+\lambda)+6 \hat{j}-2 \hat{k}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}\)
⇒ \((\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}) \cdot\left(\frac{\hat{\mathrm{i}}(2+\lambda)+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}\right)=1\)
(because \(\vec{\mathrm{a}} \cdot \vec{\mathrm{d}}=\hat{1})\)
⇒ \((2+\lambda)+6-2= \pm \sqrt{(2+\lambda)^2+36+4}\)
⇒ \(6+\lambda= \pm \sqrt{4+4 \lambda+\lambda^2+40}\)
⇒ \(36+\lambda^2+12 \lambda=44+4 \lambda+\hat{\lambda}^2 \text { or } 8 \lambda=8 \Rightarrow \lambda=1\)
Unit vector along \(\vec{b}+\vec{c}\)
= \(\frac{\hat{\mathrm{i}}(2+\lambda)+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{(2+\lambda)^2+6^2+(-2)^2}}\)
= \(\frac{3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{\sqrt{9+36+4}}=\frac{3 \hat{\mathrm{i}}+6 \hat{\mathrm{j}}-2 \hat{\mathrm{k}}}{7}\) (because \lambda=1)
Question 16. Let \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}, \vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-\hat{k}\). Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{c}\) and \(\vec{b}\) and \(\vec{d}\) \(\vec{a}\) = 21.
Solution:
Given: \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}\),
⇒ \(\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\)
⇒ \(\vec{c}=3 \hat{i}+\hat{j}-k \text { and } \vec{d} \cdot \vec{a}=21\)
⇒ \(\vec{c} \times \vec{b}\)
– \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -1 \\
1 & -4 & 5
\end{array}\right|\)
– \(\hat{i}(5-4)-\hat{j}(15+1)+\hat{k}(-12-1)\)
or \(\vec{c} \times \vec{b}=\hat{i}-16 \hat{j}-13 \hat{k}\)
Now, \(\vec{d}\) is \(\perp\) to both \(\vec{c}\) and \(\vec{b}\) (given)
i.e. \(\vec{\mathrm{d}} \|(\vec{\mathrm{c}} \times \vec{\mathrm{b}})\) or \(\vec{\mathrm{d}}=\lambda \vec{\mathrm{c}} \times \vec{\mathrm{b}}\)
⇒ \(\vec{d}=\lambda(\hat{i}-16 \hat{j}-13 \hat{k})=(\lambda \hat{i}-16 \lambda \hat{j}-13 \lambda k)\)….(1)
Now, \(\vec{\mathrm{d}} \cdot \vec{\mathrm{a}}=21\)
⇒ \((\lambda \hat{\mathrm{i}}-16 \lambda \hat{\mathrm{j}}-13 \lambda \mathrm{k}) \cdot(4 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-\mathrm{k})=21\) (from (1))
or \(4 \lambda-80 \lambda+13 \lambda=21\)
⇒ \(-63 \lambda=21 \text { or } \lambda=\frac{21}{-63}=-\frac{1}{3}\)
Put this value of \(\lambda\) in equation (1), we get :
⇒ \(\vec{\mathrm{d}}=-\frac{1}{3}(\hat{\mathrm{i}}-16 \hat{\mathrm{j}}-13 \hat{\mathrm{k}})\)
⇒ \(\vec{\mathrm{d}}=-\frac{1}{3} \hat{\mathrm{i}}+\frac{16}{3} \hat{\mathrm{j}}+\frac{13}{3} \hat{\mathrm{k}}\).