Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

Simplifying complex vector expressions often relies on the vector triple product identity, which states that $(\vec{u}\times \vec{v})\times \vec{w} = (\vec{u}\cdot\vec{w})\vec{v} - (\vec{v}\cdot\vec{w})\vec{u}$. Applying this identity to the term $(\vec{a}\times \vec{b})\times (\vec{a}+2\vec{b})$, we obtain $(\vec{a}\cdot(\vec{a}+2\vec{b}))\vec{b} - (\vec{b}\cdot(\vec{a}+2\vec{b}))\vec{a}$. Given that $\vec{a}\cdot\vec{a}=1$, $\vec{b}\cdot\vec{b}=1$, and $\vec{a}\cdot\vec{b}=0$, this expression simplifies to $(1+0)\vec{b} - (0+2)\vec{a}$, which is $\vec{b}-2\vec{a}$. The final step requires finding the dot product of this result with $(2\vec{a}-\vec{b})$, specifically $(2\vec{a}-\vec{b})\cdot(\vec{b}-2\vec{a})$. Expanding the dot product gives $4\vec{a}\cdot\vec{b} - 4\vec{a}\cdot\vec{a} - \vec{b}\cdot\vec{b}$, which results in $0 - 4(1) - 1$, yielding $-5$.

Mutually orthogonal vectors imply that the dot product of any pair in the set must equal zero, which means $\vec{b} \cdot \vec{c} = 0$ and $\vec{a} \cdot \vec{c} = 0$. Calculating the dot product $\vec{b} \cdot \vec{c} = 0$ yields $2\lambda + 4 + \mu = 0$, simplifying to $2\lambda + \mu = -4$. Similarly, evaluating the dot product $\vec{a} \cdot \vec{c} = 0$ produces $\lambda - 1 + 2\mu = 0$, or $\lambda + 2\mu = 1$. Solving the resulting system of linear equations by substituting $\mu = -4 - 2\lambda$ into the second equation yields $\lambda + 2(-4 - 2\lambda) = 1$, which simplifies to $-3\lambda - 8 = 1$, leading to $\lambda = -3$. Substituting $\lambda = -3$ back into the first equation gives $2(-3) + \mu = -4$, confirming that $\mu = 2$.

The equation $\vec{a} \times \vec{b} + \vec{c} = \vec{0}$ rearranges to $\vec{b} \times \vec{a} = \vec{c}$, which implies that $\vec{c}$ is perpendicular to $\vec{b}$, or mathematically $\vec{b} \cdot \vec{c} = 0$. Representing $\vec{b}$ as $b_1 \hat{i} + b_2 \hat{j} + b_3 \hat{k}$, the dot product with $\vec{c} = \hat{i} - \hat{j} - \hat{k}$ simplifies to $b_1 - b_2 - b_3 = 0$, which can be written as $b_1 = b_2 + b_3$. We are also given $\vec{a} \cdot \vec{b} = 3$ with $\vec{a} = \hat{j} - \hat{k}$, yielding the relation $b_2 - b_3 = 3$, or $b_2 = 3 + b_3$. Substituting this expression for $b_2$ into the equation for $b_1$ gives $b_1 = (3 + b_3) + b_3 = 3 + 2b_3$. Combining these into the general vector form $\vec{b} = (3 + 2b_3) \hat{i} + (3 + b_3) \hat{j} + b_3 \hat{k}$ allows us to evaluate the options, and setting $b_3 = -2$ produces the vector $-\hat{i} + \hat{j} - 2 \hat{k}$.

Since the vectors $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar, their scalar triple product is non-zero, allowing the original equality to be simplified to the homogeneous quadratic relation $3p^2 - pq + 2q^2 = 0$. To determine the possible real values for $p$ and $q$, we can rearrange this equation by completing the square, which yields $2p^2 + (p - \frac{q}{2})^2 + \frac{7}{4}q^2 = 0$. Because each squared term must be non-negative, the only way for their sum to equal zero is if each term is individually zero, which forces both $p = 0$ and $q = 0$. Consequently, there is exactly one solution pair $(p, q)$ that satisfies the condition.

Any vector $\vec{a}$ residing in the plane of two vectors $\vec{b}$ and $\vec{c}$ can be expressed as a linear combination $\vec{a} = \vec{b} + \lambda \vec{c}$ for some scalar $\lambda$. Substituting the given definitions, we write $\alpha \hat{i} + 2 \hat{j} + \beta \hat{k} = (\hat{i} + \hat{j}) + \lambda(\hat{j} + \hat{k})$, which simplifies to $\hat{i} + (1 + \lambda)\hat{j} + \lambda \hat{k}$. By equating the corresponding components on both sides, we identify $\alpha = 1$ from the $\hat{i}$ terms, $2 = 1 + \lambda$ which implies $\lambda = 1$ from the $\hat{j}$ terms, and $\beta = \lambda = 1$ from the $\hat{k}$ terms. Consequently, the values that satisfy these conditions are $\alpha = 1$ and $\beta = 1$.

Two vectors are considered collinear when one is a scalar multiple of the other, indicating they lie along the same line. To determine the orientation between $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{c}$, express both in terms of the common vector $\overset{\rightarrow}{b}$. Given $\overset{\rightarrow}{c}=-7 \overset{\rightarrow}{b}$, we can isolate $\overset{\rightarrow}{b}$ to get $\overset{\rightarrow}{b}=-\frac{1}{7} \overset{\rightarrow}{c}$. Substituting this expression into $\overset{\rightarrow}{a}=8 \overset{\rightarrow}{b}$ results in $\overset{\rightarrow}{a}=-\frac{8}{7} \overset{\rightarrow}{c}$. Because the scalar constant connecting these two vectors is negative, they point in diametrically opposite directions, which corresponds to an angle of $\pi$.

When three vectors lie in the same plane, they are coplanar, which implies that their scalar triple product is zero. We express this condition by setting the determinant of the matrix formed by the vector components equal to zero.

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$

Expanding this determinant along the first row gives $1(1 - 2(x-2)) - 1(-1 - 2x) + 1(x-2+x) = 0$. Simplifying the terms results in $(5 - 2x) + (1 + 2x) + (2x - 2) = 0$, which further reduces to $2x + 4 = 0$. Solving this equation for $x$ leads directly to $x = -2$.

The magnitude of the cross product involving $2\hat{u}$ and $3\hat{v}$ is determined by the expression $6 |\hat{u}| |\hat{v}| \sin\theta$. Given that both $\hat{u}$ and $\hat{v}$ are unit vectors, this simplifies to $6 \sin\theta$. Requiring this magnitude to equal one results in the equation $6 \sin\theta = 1$, which means $\sin\theta = \frac{1}{6}$. Since the sine function corresponds to exactly one angle in the range of acute angles, there is exactly one such $\theta$.

For a triangle to have a right angle at vertex $C$, the vectors representing the two sides meeting at that vertex, $\vec{CA}$ and $\vec{CB}$, must be perpendicular. This geometric condition implies that the dot product of these two vectors must be equal to zero.

Calculating the displacement vectors by subtracting the position vector of $C$ from those of $A$ and $B$ respectively, we determine that $\vec{CA} = (2-a)\hat{i} + 2\hat{j}$ and $\vec{CB} = (1-a)\hat{i} - 6\hat{k}$. Applying the condition $\vec{CA} \cdot \vec{CB} = 0$ results in the following equation:

$(2-a)(1-a) = 0$

Solving this quadratic equation directly provides the values for $a$ as $2$ and $1$.

The vector triple product identity enables us to express $(\vec{a} \times \vec{b}) \times \vec{c}$ as $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$ and $\vec{a} \times (\vec{b} \times \vec{c})$ as $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Equating these two expansions, we can cancel the common term $(\vec{a} \cdot \vec{c})\vec{b}$ from both sides of the equation. This simplification leaves us with $-(\vec{b} \cdot \vec{c})\vec{a} = -(\vec{a} \cdot \vec{b})\vec{c}$, which can be rewritten as $(\vec{b} \cdot \vec{c})\vec{a} = (\vec{a} \cdot \vec{b})\vec{c}$. Because the dot products $(\vec{b} \cdot \vec{c})$ and $(\vec{a} \cdot \vec{b})$ are scalar quantities, this final equation demonstrates that vector $\vec{a}$ is a scalar multiple of vector $\vec{c}$. Therefore, $\vec{a}$ and $\vec{c}$ must be parallel.

The scalar triple product exhibits linearity, which allows for the extraction of scalar multipliers and the distribution of vectors across additive terms within the brackets. Factoring the scalars $\lambda$, $\lambda^2$, and $\lambda$ from the left side of the equation yields $\lambda^4 [\vec{a}+\vec{b} \quad \vec{b} \quad \vec{c}]$. Since a scalar triple product vanishes whenever any two vectors are identical, the term $[\vec{a}+\vec{b} \quad \vec{b} \quad \vec{c}]$ simplifies to $[\vec{a} \quad \vec{b} \quad \vec{c}]$. Expanding the right side $[\vec{a} \quad \vec{b}+\vec{c} \quad \vec{b}]$ gives $[\vec{a} \quad \vec{b} \quad \vec{b}] + [\vec{a} \quad \vec{c} \quad \vec{b}]$, where the first term is zero and the second simplifies to $-[\vec{a} \quad \vec{b} \quad \vec{c}]$ due to the anti-symmetry property of swapping two vectors. Equating these results leads to $\lambda^4 [\vec{a} \quad \vec{b} \quad \vec{c}] = -[\vec{a} \quad \vec{b} \quad \vec{c}]$, which requires $\lambda^4 = -1$ because the non-coplanarity of the vectors ensures the scalar triple product is non-zero. Since no real number raised to the fourth power can be negative, there are no real values of $\lambda$ that satisfy the condition.

The relationship between vectors in this scenario is governed by the triangle law of vector addition, which allows us to decompose the vectors $\overrightarrow{PA}$ and $\overrightarrow{PB}$ through the midpoint $C$ as $\overrightarrow{PA} = \overrightarrow{PC} + \overrightarrow{CA}$ and $\overrightarrow{PB} = \overrightarrow{PC} + \overrightarrow{CB}$. Because $C$ is the midpoint of the segment $AB$, the vectors $\overrightarrow{CA}$ and $\overrightarrow{CB}$ are additive inverses, meaning $\overrightarrow{CA} = -\overrightarrow{CB}$ and thus $\overrightarrow{CA} + \overrightarrow{CB} = \overrightarrow{0}$. Adding these two expressions together yields $\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC} + (\overrightarrow{CA} + \overrightarrow{CB})$, which simplifies directly to $\overrightarrow{PA} + \overrightarrow{PB} = 2\overrightarrow{PC}$ since the remaining term is zero.

The scalar triple product of three vectors, denoted as $[\vec{a}, \vec{b}, \vec{c}]$, represents the volume of the parallelepiped spanned by the vectors and is calculated by evaluating the determinant of the matrix formed by their components. Setting up the determinant with the given vectors $\vec{a} = \hat{i} - \hat{k}$, $\vec{b} = x\hat{i} + \hat{j} + (1-x)\hat{k}$, and $\vec{c} = y\hat{i} + x\hat{j} + (1+x-y)\hat{k}$ provides the matrix:
$[\vec{a}, \vec{b}, \vec{c}] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1-x \\ y & x & 1+x-y \end{vmatrix}$
Expanding this determinant along the first row gives $1(1+x-y - x(1-x)) - 0 + (-1)(x^2 - y)$, which simplifies to $(1 + x - y - x + x^2) - x^2 + y$. Combining these terms, the $x$, $x^2$, and $y$ terms all cancel out, leaving a constant value of $1$. Since the final result is a constant, the scalar triple product is independent of both $x$ and $y$.

When three vectors are coplanar, they reside within a single plane, which requires their scalar triple product to be zero. This condition is mathematically equivalent to setting the determinant of the matrix formed by the vector components equal to zero. The determinant for the vectors $a\hat{i} + a\hat{j} + c\hat{k}$, $\hat{i} + \hat{k}$, and $c\hat{i} + c\hat{j} + b\hat{k}$ is:

$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$

Expanding this expression results in $a(0 - c) - a(b - c) + c(c - 0) = 0$, which simplifies to $-ac - ab + ac + c^2 = 0$. This leads to $c^2 - ab = 0$, or $c^2 = ab$, proving that $c = \sqrt{ab}$. Consequently, $c$ is the geometric mean of $a$ and $b$.

Expressing a vector as $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ allows for the evaluation of its cross products with the coordinate unit vectors, resulting in $\vec{a} \times \hat{i} = z\hat{j} - y\hat{k}$, $\vec{a} \times \hat{j} = x\hat{k} - z\hat{i}$, and $\vec{a} \times \hat{k} = y\hat{i} - x\hat{j}$. Determining the squared magnitude for each of these three vectors gives $y^2 + z^2$, $x^2 + z^2$, and $x^2 + y^2$ respectively. Summing these values yields $2(x^2 + y^2 + z^2)$, which simplifies to $2a^2$ based on the relationship $a^2 = x^2 + y^2 + z^2$.

Two non-zero vectors are defined as collinear if one is a scalar multiple of the other, which allows us to express the given collinearity conditions as $\vec{a} + 2\vec{b} = t\vec{c}$ and $\vec{b} + 3\vec{c} = s\vec{a}$ for some scalar constants $t$ and $s$. When evaluating the target expression $\vec{a} + 2\vec{b} + 6\vec{c}$, we can directly substitute the first condition by replacing $\vec{a} + 2\vec{b}$ with $t\vec{c}$. This substitution transforms the expression into $t\vec{c} + 6\vec{c}$, which factors to $(t + 6)\vec{c}$. Given that $t$ is a constant, the sum $(t + 6)$ represents another scalar value, which the problem identifies as $\lambda$, confirming the final result is $\lambda\vec{c}$.

$\;\vec{A} \times \vec{B}=\vec{B} \times \vec{A}$ $\;\vec{A} \times \vec{B}-\vec{B} \times \vec{A}=0$ $\;\Rightarrow \vec{A} \times \vec{B}+\vec{A} \times \vec{B}=0$ $\; \therefore \vec{A} \times \vec{B}=0$ $\;\Rightarrow A B \sin \;\theta =0$ $\;\theta =0, \pi , 2 \pi . . . . . . .$from the given options, $\theta =\pi$

The magnitude of a vector expression can be determined by evaluating the square of its norm using the distributive property of the dot product, expressed by the identity $|\vec{u}-\vec{v}+\vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 - 2(\vec{u} \cdot \vec{v}) + 2(\vec{u} \cdot \vec{w}) - 2(\vec{v} \cdot \vec{w})$. Because the scalar projections of $\vec{v}$ and $\vec{w}$ along $\vec{u}$ are equal, the dot products $\vec{u} \cdot \vec{v}$ and $\vec{u} \cdot \vec{w}$ are equivalent, causing the terms $-2(\vec{u} \cdot \vec{v})$ and $2(\vec{u} \cdot \vec{w})$ to cancel each other out. Additionally, since $\vec{v}$ and $\vec{w}$ are perpendicular, their dot product $\vec{v} \cdot \vec{w}$ is zero, which eliminates the final term in the expansion. Substituting the given magnitudes $|\vec{u}|=1$, $|\vec{v}|=2$, and $|\vec{w}|=3$ into the simplified expression results in $|\vec{u}-\vec{v}+\vec{w}|^2 = 1^2 + 2^2 + 3^2$, which equals 14, meaning the final magnitude is $\sqrt{14}$.

Work done on a particle by multiple constant forces is determined by the dot product of the resultant force vector and the displacement vector. We first find the net force $\vec{F}$ by summing the individual force vectors, yielding $\vec{F} = (4 \hat{i} + \hat{j} - 3 \hat{k}) + (3 \hat{i} + \hat{j} - \hat{k}) = 7 \hat{i} + 2 \hat{j} - 4 \hat{k}$.

Next, the displacement vector $\vec{d}$ is the difference between the final position and the initial position vectors, which is calculated as $\vec{d} = (5 \hat{i} + 4 \hat{j} + \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 4 \hat{i} + 2 \hat{j} - 2 \hat{k}$. Finally, the work done is the scalar product of these two vectors, $W = \vec{F} \cdot \vec{d} = (7 \hat{i} + 2 \hat{j} - 4 \hat{k}) \cdot (4 \hat{i} + 2 \hat{j} - 2 \hat{k})$, which simplifies to $28 + 4 + 8$ for a result of 40.

Three vectors are non-coplanar if their scalar triple product is non-zero, which is equivalent to requiring the determinant of the matrix formed by their coefficients to be non-zero. Constructing this determinant using the components of the given vectors $\vec{a}+2\vec{b}+3\vec{c}$, $\lambda\vec{b}+4\vec{c}$, and $(2\lambda-1)\vec{c}$ relative to the basis $\vec{a}, \vec{b}, \vec{c}$ yields:
$\begin{vmatrix} 1 & 2 & 3 \\ 0 & \lambda & 4 \\ 0 & 0 & 2\lambda-1 \end{vmatrix}$
Since this is an upper triangular matrix, its determinant is simply the product of the diagonal elements, resulting in the expression $\lambda(2\lambda-1)$. Setting this product equal to zero identifies the values of $\lambda$ for which the vectors are coplanar, specifically $\lambda = 0$ and $\lambda = \frac{1}{2}$. Therefore, the vectors remain non-coplanar for every real number $\lambda$ except these two values.