Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

Work is the scalar product of the net force acting on a particle and its displacement. The resultant force is determined by adding the given force vectors to obtain $\vec{F} = 7\hat{i}+2\hat{j}-4\hat{k}$. Using the specified displacement vector $\vec{d} = 4\hat{i}+2\hat{j}-2\hat{k}$, the total work done is the dot product $W = \vec{F} \cdot \vec{d} = (7\hat{i}+2\hat{j}-4\hat{k}) \cdot (4\hat{i}+2\hat{j}-2\hat{k})$. This calculation results in $28 + 4 + 8 = 40$ units.

Expanding the vector triple product $(\vec{a} \times \vec{b}) \times \vec{c}$ using the fundamental identity yields the expression $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$. Equating this result to $\frac{1}{3} |\vec{b}| |\vec{c}| \vec{a}$ allows for a comparison of the coefficients associated with the non-collinear vectors $\vec{a}$ and $\vec{b}$. This comparison reveals that $\vec{a} \cdot \vec{c} = 0$ and the negative scalar product $-(\vec{b} \cdot \vec{c})$ must be equal to $\frac{1}{3} |\vec{b}| |\vec{c}|$. Substituting the geometric definition of the dot product $\vec{b} \cdot \vec{c} = |\vec{b}| |\vec{c}| \cos \theta$ into this equality leads to $\cos \theta = -\frac{1}{3}$. Applying the identity $\sin \theta = \sqrt{1 - \cos^2 \theta}$ results in $\sqrt{1 - (-\frac{1}{3})^2}$, which simplifies to $\frac{2\sqrt{2}}{3}$.

The distributive property of vector products allows us to expand the expression by first computing the cross product of the second and third vectors, which simplifies to $\vec{u} \times \vec{v} - \vec{u} \times \vec{w} + \vec{v} \times \vec{w}$ because any vector crossed with itself equals the zero vector. Taking the dot product of the first vector $\vec{u} + \vec{v} - \vec{w}$ with this expansion yields nine terms, most of which vanish because the scalar triple product is zero whenever any two of the vectors are identical. By discarding these zero terms, we are left with the non-zero contributions $\vec{u} \cdot (\vec{v} \times \vec{w}) - \vec{v} \cdot (\vec{u} \times \vec{w}) - \vec{w} \cdot (\vec{u} \times \vec{v})$. Since each of these non-zero terms is equivalent to the scalar triple product $\vec{u} \cdot \vec{v} \times \vec{w}$ through the cyclic property, the entire expression simplifies directly to $\vec{u} \cdot \vec{v} \times \vec{w}$.

The core principle involves expanding the square of a vector sum using the dot product, which relates the magnitudes of the individual vectors to the sum of their pairwise dot products. Given the expression $\vec{a}+\vec{b}+\vec{c}=0$, taking the dot product of this sum with itself yields $(\vec{a}+\vec{b}+\vec{c}) \cdot (\vec{a}+\vec{b}+\vec{c}) = 0$.

Expanding this product using the distributive property leads to the equation:
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$
Substituting the squared magnitudes 1, 4, and 9 for $\vec{a}$, $\vec{b}$, and $\vec{c}$ respectively, the equation becomes $1 + 4 + 9 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$. Simplifying this results in $14 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) = 0$. Isolating the desired sum of dot products by subtracting 14 and dividing by 2 leaves -7.

The dihedral angle between two planes in space is equal to the angle between their respective normal vectors. To determine the normal vector for each face of the tetrahedron, we calculate the cross product of two edge vectors lying on that face.

For the face $OAB$, the edge vectors are $\overrightarrow{OA} = (1, 2, 1)$ and $\overrightarrow{OB} = (2, 1, 3)$. Calculating their cross product $\overrightarrow{OA} \times \overrightarrow{OB}$ yields the normal vector $\vec{n}_1 = 5\hat{i} - \hat{j} - 3\hat{k}$. For the face $ABC$, we identify the edge vectors $\overrightarrow{AB} = B - A = (1, -1, 2)$ and $\overrightarrow{AC} = C - A = (-2, -1, 1)$. Their cross product $\overrightarrow{AB} \times \overrightarrow{AC}$ results in the second normal vector $\vec{n}_2 = \hat{i} - 5\hat{j} - 3\hat{k}$.

With both normal vectors determined, the angle $\theta$ between the faces is found using the dot product formula $\cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1| |\vec{n}_2|}$. The dot product of the two vectors is $5(1) + (-1)(-5) + (-3)(-3) = 19$, and each vector has a magnitude of $\sqrt{5^2 + (-1)^2 + (-3)^2} = \sqrt{35}$. Substituting these values, we get $\cos \theta = \frac{19}{\sqrt{35} \sqrt{35}}$, which simplifies to $\cos \theta = \frac{19}{35}$. Thus, the angle is $\theta = \cos^{-1} \left( \frac{19}{35} \right)$.

A unit vector perpendicular to two given vectors is always parallel to their cross product. Computing the cross product of $\vec{u} = \hat{i} + \hat{j}$ and $\vec{v} = \hat{i} - \hat{j}$ produces the vector

. Since the magnitude of this result is 2, the unit vector $\hat{n}$ is obtained by dividing by its magnitude, resulting in $\hat{n} = \pm \hat{k}$. Applying the dot product with $\vec{w} = \hat{i} + 2\hat{j} + 3\hat{k}$ yields $\vec{w} \cdot (\pm \hat{k}) = \pm 3$, and taking the absolute value gives 3.

Calculating the side lengths of the quadrilateral using the distance formula is the primary method to identify its geometry. The distance between points $A$ and $B$ is $AB = \sqrt{(7-1)^{2+}(-4+6)^{2+}(7-10)^2} = \sqrt{36+4+9} = 7$, and the distance between points $B$ and $C$ is $BC = \sqrt{(1-(-1))^{2+}(-6-(-3))^{2+}(10-4)^2} = \sqrt{4+9+36} = 7$. For the remaining sides, the calculation yields $CD = \sqrt{(-1-5)^{2+}(-3-(-1))^{2+}(4-5)^2} = \sqrt{36+4+1} = \sqrt{41}$ and $DA = \sqrt{(5-7)^{2+}(-1-(-4))^{2+}(5-7)^2} = \sqrt{4+9+4} = \sqrt{17}$. As the sides $7, 7, \sqrt{41}, \text{ and } \sqrt{17}$ are not congruent in a way that satisfies the definitions of a parallelogram, rhombus, or square, none of the given classifications apply to the figure.

In any triangle $ABC$ where the sides originating from vertex $A$ are represented by vectors $\vec{AB}$ and $\vec{AC}$, the median $\vec{AD}$ directed towards the midpoint of $BC$ is simply the average of these two side vectors. Because the vector to the midpoint of a side is the arithmetic mean of the vectors to the endpoints, we find the median vector using:
$\vec{AD} = \frac{\vec{AB} + \vec{AC}}{2}$

Combining the given vectors $\vec{AB} = 3\hat{i} + 4\hat{k}$ and $\vec{AC} = 5\hat{i} - 2\hat{j} + 4\hat{k}$, we calculate their sum as $(3+5)\hat{i} + (0-2)\hat{j} + (4+4)\hat{k}$, which simplifies to $8\hat{i} - 2\hat{j} + 8\hat{k}$. Dividing this vector by two results in the median vector $\vec{AD} = 4\hat{i} - \hat{j} + 4\hat{k}$. Finally, the length is determined by the magnitude formula $|\vec{AD}| = \sqrt{4^2 + (-1)^2 + 4^2}$, which results in $\sqrt{16 + 1 + 16}$ or $\sqrt{33}$.

Vector addition principles dictate that if the sum of three vectors is zero, the magnitude of the sum of two vectors is equal to the magnitude of the remaining vector. Squaring the expression $\overset{\rightarrow}{b}+\overset{\rightarrow}{c}=-\overset{\rightarrow}{a}$ yields the relationship $|\overset{\rightarrow}{b}|^2 + |\overset{\rightarrow}{c}|^2 + 2|\overset{\rightarrow}{b}||\overset{\rightarrow}{c}| \cos \theta = |\overset{\rightarrow}{a}|^2$. Inserting the known magnitudes of 5, 3, and 7 results in $5^2 + 3^2 + 2(5)(3) \cos \theta = 7^2$. Simplifying the arithmetic gives $25 + 9 + 30 \cos \theta = 49$, which further reduces to $30 \cos \theta = 15$. Isolating the cosine function provides $\cos \theta = \frac{1}{2}$, indicating that the angle between the two vectors is 60°.

The magnitude of the cross product of two vectors is defined by the relationship $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$. Squaring both sides of this identity yields the formula
$(\vec{a} \times \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$
Substituting the given magnitudes of 4 and 2 along with the angle $\frac{\pi}{6}$ into this equation results in $16 \times 4 \times \sin^2\left(\frac{\pi}{6}\right)$. Since $\sin\left(\frac{\pi}{6}\right)$ evaluates to $\frac{1}{2}$, its square is $\frac{1}{4}$, making the final calculation $16 \times 4 \times \frac{1}{4}$, which simplifies to 16.

To determine the ratio of these vector magnitudes, first calculate the individual lengths of $\vec{a}$ and $\vec{b}$, then find the magnitude of their cross product $\vec{c}$. The magnitude $|\vec{a}|$ for the vector $3\hat{i}-5\hat{j}$ is calculated as $\sqrt{3^{2+}(-5)^2} = \sqrt{34}$, while the magnitude $|\vec{b}|$ for $6\hat{i}+3\hat{j}$ is $\sqrt{6^{2+}3^2} = \sqrt{45}$. Computing the cross product $\vec{c} = (3\hat{i}-5\hat{j}) \times (6\hat{i}+3\hat{j})$ results in $9(\hat{i} \times \hat{j}) - 30(\hat{j} \times \hat{i})$, which simplifies to $9\hat{k} - 30(-\hat{k}) = 39\hat{k}$, yielding a magnitude $|\vec{c}| = 39$. Consequently, the final ratio $|\vec{a}|:|\vec{b}|:|\vec{c}|$ is $\sqrt{34}:\sqrt{45}:39$.

Assigning $\vec{r} = \vec{a} + \vec{b} + \vec{c}$ and calculating the cross product $\vec{a} \times \vec{r}$ results in $\vec{a} \times \vec{a} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$. Given that $\vec{a} \times \vec{a} = \vec{0}$ and $\vec{a} \times \vec{c} = -\vec{c} \times \vec{a}$, this expression simplifies to $\vec{a} \times \vec{b} - \vec{c} \times \vec{a}$, which is equal to $\vec{0}$ according to the provided condition $\vec{a} \times \vec{b} = \vec{c} \times \vec{a}$. Since this same pattern applies to $\vec{b} \times \vec{r} = \vec{0}$ and $\vec{c} \times \vec{r} = \vec{0}$, we can conclude that $\vec{r} = \vec{0}$.

The scalar triple product of three vectors, denoted as $[\vec{a} \vec{b} \vec{c}]$, corresponds to the volume of the parallelepiped formed by these vectors. A fundamental property in vector algebra establishes that the scalar triple product of the cross products of three vectors is equal to the square of the original scalar triple product. This relationship is expressed as:
$[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = [\vec{a} \vec{b} \vec{c}]^2$

Given that $[\vec{a} \vec{b} \vec{c}] = 4$, we can directly substitute this value into the identity. Squaring the value yields $4^2$, resulting in 16.

The arrangement of vectors in a right-handed system, specifically in the sequence $\vec{a}, \vec{c}, \vec{b}$, dictates that the vector $\vec{c}$ is defined by the cross product $\vec{b} \times \vec{a}$. To calculate this, we set up a determinant with the unit vectors $\hat{i}, \hat{j}, \hat{k}$ in the first row, the components of $\vec{b}$ which are $(0, 1, 0)$ in the second row, and the components of $\vec{a}$ which are $(x, y, z)$ in the bottom row. Calculating the determinant

results in the expression $\hat{i}(1 \cdot z - 0 \cdot y) - \hat{j}(0 \cdot z - 0 \cdot x) + \hat{k}(0 \cdot y - 1 \cdot x)$, which simplifies to $z \hat{i} - x \hat{k}$.

Squaring the sum of three vectors allows us to isolate pairwise dot products, leveraging the fact that the dot product of a vector with itself equals its magnitude squared. Expanding the identity $(\vec{a} + \vec{b} + \vec{c}) \cdot (\vec{a} + \vec{b} + \vec{c}) = 0$ yields the expression $|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$, which simplifies to $50 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$. This equality reveals that the \sum of the dot products is $-25$. Taking the absolute value of this \sum gives the final result of $25$.

The projection of vector $\vec{a}$ onto vector $\vec{c}$ is determined by the formula $\frac{\vec{a} \cdot \vec{c}}{|\vec{c}|}$. Calculating the dot product gives $(\alpha)(1) + (3)(2) + (-1)(-2)$, which simplifies to $\alpha + 6 + 2 = \alpha + 8$, while the magnitude of $\vec{c}$ is $\sqrt{1^2 + 2^2 + (-2)^2} = 3$. By equating the projection $\frac{\alpha + 8}{3}$ to the given value of $\frac{10}{3}$, we find $\alpha = 2$.

To determine $\beta$, we compute the cross product $\vec{b} \times \vec{c}$ using the determinant method:
$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -\beta & 4 \\ 1 & 2 & -2 \end{vmatrix}$
Expanding this yields $(2\beta - 8)\hat{i} + 10\hat{j} + (6 + \beta)\hat{k}$. Comparing this result to the given vector $-6 \hat{i} + 10 \hat{j} + 7 \hat{k}$ allows us to equate the corresponding components. Setting the $\hat{i}$ coefficient equal gives $2\beta - 8 = -6$, which simplifies to $2\beta = 2$, meaning $\beta = 1$. Verifying this with the $\hat{k}$ component, $6 + 1 = 7$, confirms the value, and adding the two variables provides $\alpha + \beta = 2 + 1 = 3$.

The vector $\vec{b}$ represents a 90 degree rotation of $\vec{a}$ within the plane formed by $\vec{a}$ and the $y$-axis unit vector $\hat{j}$. Any vector in this plane can be expressed as a linear combination of $\vec{a}$ and $\vec{a} \times (\vec{a} \times \hat{j})$, ensuring it remains perpendicular to $\vec{a}$ and maintains the same magnitude $\sqrt{6}$. Calculating the cross product $\vec{a} \times (\vec{a} \times \hat{j})$ yields the vector $-2\hat{i}-2\hat{j}+2\hat{k}$, which serves as the directional basis for $\vec{b}$. By applying the magnitude requirement and ensuring $\vec{b}$ forms an acute angle with the $y$-axis, we determine $\vec{b} = \sqrt{2}\hat{i}+\sqrt{2}\hat{j}-\sqrt{2}\hat{k}$.

The linear combination $3\vec{a} + \sqrt{2}\vec{b}$ simplifies to $-3\hat{i} + 6\hat{j} + 3\hat{k} + 2\hat{i} + 2\hat{j} - 2\hat{k}$, resulting in the vector $-\hat{i} + 8\hat{j} + \hat{k}$. The projection of this resulting vector onto $\vec{c} = 5\hat{i} + 4\hat{j} + 3\hat{k}$ is found by calculating the dot product and dividing by the magnitude of $\vec{c}$. The dot product is 30 and the magnitude of $\vec{c}$ is $\sqrt{50}$, which is equal to $5\sqrt{2}$. Dividing these values results in $3\sqrt{2}$.

Since the vector $\vec{c}$ is coplanar with $\vec{a}=-\hat{i}+\hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$, it can be expressed as $\vec{c} = x\vec{a} + y\vec{b}$ for some scalars $x$ and $y$. Using the vector triple product property $\vec{b} \times (\vec{a} \times \vec{b}) = (\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b}$, we calculate the terms as $\vec{b} \cdot \vec{b} = 6$ and $\vec{b} \cdot \vec{a} = -2$, giving $\vec{b} \times (\vec{a} \times \vec{b}) = 6\vec{a} + 2\vec{b}$. The first condition simplifies as:
$\vec{c} \cdot [(\vec{c}+\vec{b}) \times (\vec{a} \times \vec{b})] = \vec{c} \cdot [\vec{c} \times (\vec{a} \times \vec{b}) + \vec{b} \times (\vec{a} \times \vec{b})] = \vec{c} \cdot (6\vec{a} + 2\vec{b}) = -42$
Expanding this, $6(\vec{c} \cdot \vec{a}) + 2(\vec{c} \cdot \vec{b}) = -42$, which simplifies to $3(\vec{c} \cdot \vec{a}) + (\vec{c} \cdot \vec{b}) = -21$. Substituting $\vec{c} \cdot \vec{a} = 3x - 2y$ and $\vec{c} \cdot \vec{b} = -2x + 6y$ yields $3(3x-2y) + (-2x+6y) = 7x = -21$, so $x = -3$.

The second condition $(\vec{c} \times (\vec{a}-\vec{b})) \cdot \hat{k} = 3$ can be expressed as:
$(\vec{c} \times (\vec{a}-\vec{b})) \cdot \hat{k} = -(x+y)(\vec{a} \times \vec{b}) \cdot \hat{k} = 3$
Using $\vec{a} \times \vec{b} = -2\hat{i} + \hat{j} - 3\hat{k}$, the condition becomes $-(x+y)(-3) = 3$, which implies $x+y=1$. With $x=-3$, we find $y=4$. The vector $\vec{c}$ is thus $-3\vec{a} + 4\vec{b} = -3(-\hat{i}+\hat{j}+\hat{k}) + 4(2\hat{i}+\hat{j}-\hat{k}) = 11\hat{i} + \hat{j} - 7\hat{k}$. Finally, the square of the magnitude is:
$|\vec{c}|^2 = 11^2 + 1^2 + (-7)^2 = 121 + 1 + 49 = 171$
(Note: Given the discrepancy between the derived value and the provided options, the method above is the standard mathematical procedure for this problem type.)

The direction of the line in question is uniquely determined by the cross product of the direction vectors of the two given lines, as it must be orthogonal to both. Given the direction vectors $2\hat{i}+3\hat{j}+5\hat{k}$ and $-\hat{i}+3\hat{j}+2\hat{k}$, we calculate their cross product using the determinant

, which results in $-9\hat{i}-9\hat{j}+9\hat{k}$. We can simplify this direction to $\hat{i}+\hat{j}-\hat{k}$, so the line passing through $P(5, -4, 3)$ with this direction is defined by the equation $\vec{r} = (5\hat{i}-4\hat{j}+3\hat{k}) + \lambda(\hat{i}+\hat{j}-\hat{k})$.

A general point $M$ on this line has coordinates $(5+\lambda, -4+\lambda, 3-\lambda)$. The vector $\vec{QM}$ from the point $Q(0, 2, -2)$ to $M$ is $(5+\lambda)\hat{i} + (\lambda-6)\hat{j} + (5-\lambda)\hat{k}$. Since the distance is measured along the perpendicular, the dot product of $\vec{QM}$ and the line's direction vector $\hat{i}+\hat{j}-\hat{k}$ must equal zero. Evaluating the equation $(5+\lambda)(1) + (\lambda-6)(1) + (5-\lambda)(-1) = 0$ simplifies to $5+\lambda+\lambda-6-5+\lambda = 0$, which results in $3\lambda - 6 = 0$, or $\lambda = 2$. Substituting $\lambda = 2$ into the expression for $\vec{QM}$ yields $7\hat{i}-4\hat{j}+3\hat{k}$, and its magnitude $\sqrt{7^{2+}(-4)^{2+}3^2} = \sqrt{74}$ provides the required distance.

The vector triple product expansion $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ allows us to relate the given expression $\vec{b} - \frac{\vec{c}}{2}$ to the scalar values $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = \frac{1}{2}$. Given the condition $\vec{b} \cdot \vec{d} = \vec{a} \cdot \vec{b}$, it follows that $\vec{b} \cdot \vec{d} = \frac{1}{2}$.

The required dot product $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d})$ is equivalent to the scalar triple product $\vec{a} \cdot (\vec{b} \times (\vec{c} \times \vec{d}))$. Applying the triple product expansion to the nested term $\vec{b} \times (\vec{c} \times \vec{d})$ yields $(\vec{b} \cdot \vec{d})\vec{c} - (\vec{b} \cdot \vec{c})\vec{d}$. Since $\vec{b} \cdot \vec{c} = 0$, the expression simplifies further to $(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})$. Substituting the previously identified values, we calculate $\frac{1}{2} \times \frac{1}{2}$, which results in $\frac{1}{4}$.