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Q13:
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 $\lambda$.
Solution :
Step 1: Defining the Given Vectors
Let the given vectors be defined as follows:
- $\vec{a} = \hat{i} + \hat{j} + \hat{k}$
- $\vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}$
- $\vec{c} = \lambda\hat{i} + 2\hat{j} + 3\hat{k}$
Step 2: Calculating the Sum Vector
Let $\vec{s}$ be the sum of vectors $\vec{b}$ and $\vec{c}$ [Vector Addition Law].
$\vec{s} = \vec{b} + \vec{c}$
$\vec{s} = (2 + \lambda)\hat{i} + (4 + 2)\hat{j} + (-5 + 3)\hat{k}$
$\vec{s} = (2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}$
Step 3: Determining the Unit Vector along $\vec{s}$
The unit vector $\hat{s}$ is defined as $\frac{\vec{s}}{|\vec{s}|}$ [Definition of a Unit Vector].
$|\vec{s}| = \sqrt{(2 + \lambda)^2 + (6)^2 + (-2)^2}$
$|\vec{s}| = \sqrt{4 + 4\lambda + \lambda^2 + 36 + 4} = \sqrt{\lambda^2 + 4\lambda + 44}$
$\hat{s} = \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}}$
Step 4: Formulating the Scalar Product Equation
The problem states that the scalar product of $\vec{a}$ and $\hat{s}$ is equal to $1$.
$\vec{a} \cdot \hat{s} = 1$
$(\hat{i} + \hat{j} + \hat{k}) \cdot \left[ \frac{(2 + \lambda)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{\lambda^2 + 4\lambda + 44}} \right] = 1$
Step 5: Algebraic Manipulation and Solving for $\lambda$
Perform the dot product of the components [Properties of the Scalar Product]:
$\frac{1(2 + \lambda) + 1(6) + 1(-2)}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1$
$\frac{2 + \lambda + 6 - 2}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1$
$\frac{\lambda + 6}{\sqrt{\lambda^2 + 4\lambda + 44}} = 1$
$\lambda + 6 = \sqrt{\lambda^2 + 4\lambda + 44}$
Square both sides to eliminate the radical:
$(\lambda + 6)^2 = \lambda^2 + 4\lambda + 44$
$\lambda^2 + 12\lambda + 36 = \lambda^2 + 4\lambda + 44$
Subtract $\lambda^2$ from both sides and simplify:
$12\lambda - 4\lambda = 44 - 36$
$8\lambda = 8$
$\lambda = 1$
Final Solution: \lambda = 1
More Questions from Class 12 Mathematics Vector Algebra Miscellaneous Exercise on Chapter 10
- Q1: Write down a unit vector in XY-plane, making an angle of $30^{\circ}$ with the positive direction of x-axis.
- Q2: Find the scalar components and magnitude of the vector joining the points $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$.
- Q3: A girl walks 4 km towards west, then she walks 3 km in a direction $30^{\circ}$ east of north and stops. Determine the girl’s displacement from her initial point of departure.
- Q4: If $\vec{a} = \vec{b} + \vec{c}$, then is it true that $|\vec{a}| = |\vec{b}| + |\vec{c}|$? Justify your answer.
- Q5: Find the value of $x$ for which $x(\hat{i} + \hat{j} + \hat{k})$ is a unit vector.
- Q6: 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 $\vec{b} = \hat{i} - 2\hat{j} + \hat{k}$.
- Q7: If $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{c} = \hat{i} - 2\hat{j} + \hat{k}$, find a unit vector parallel to the vector $2\vec{a} - \vec{b} + 3\vec{c}$.
- Q8: 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.
- Q9: 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.
- Q10: 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.
- Q11: Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)$.
- Q12: 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 $\vec{c} \cdot \vec{d} = 15$.
- Q14: If $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}, \vec{b}$ and $\vec{c}$.
- Q15: Prove that $(\vec{a} + \vec{b}) \cdot (\vec{a} + \vec{b}) = |\vec{a}|^2 + |\vec{b}|^2$, if and only if $\vec{a}, \vec{b}$ are perpendicular, given $\vec{a} \neq \vec{0}, \vec{b} \neq \vec{0}$.
- Q16: If $\theta$ is the angle between two vectors $\vec{a}$ and $\vec{b}$, then $\vec{a} \cdot \vec{b} \ge 0$ only when
- Q17: Let $\vec{a}$ and $\vec{b}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{a} + \vec{b}$ is a unit vector if
- Q18: The value of $\hat{i} \cdot (\hat{j} \times \hat{k}) + \hat{j} \cdot (\hat{i} \times \hat{k}) + \hat{k} \cdot (\hat{i} \times \hat{j})$ is
- Q19: If $\theta$ is the angle between any two vectors $\vec{a}$ and $\vec{b}$, then $|\vec{a} \cdot \vec{b}| = |\vec{a} \times \vec{b}|$ when $\theta$ is equal to
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