Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

The equality $\vec{d} \times \vec{b} = \vec{c} \times \vec{b}$ implies that the difference between the two vectors must be parallel to $\vec{b}$, allowing us to express the vector as $\vec{d} = \vec{c} + \lambda \vec{b}$ for some scalar $\lambda$. Substituting this expression into the condition $\vec{d} \cdot \vec{a} = 24$ gives the equation $(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 24$, which rearranges to $\lambda (\vec{b} \cdot \vec{a}) = 24 - (\vec{c} \cdot \vec{a})$. Computing the necessary dot products from the given vectors, we find $\vec{c} \cdot \vec{a} = 6$ and $\vec{b} \cdot \vec{a} = 9$, which leads to $\lambda = \frac{24 - 6}{9} = 2$. Having determined $\lambda$, we calculate $\vec{d} = (2\hat{i} - \hat{j} + 4\hat{k}) + 2(3\hat{i} - 2\hat{j} + 7\hat{k})$, resulting in $\vec{d} = 8\hat{i} - 5\hat{j} + 18\hat{k}$. The squared magnitude is the sum of the squares of the components, specifically $8^2 + (-5)^2 + 18^2 = 64 + 25 + 324$, yielding 413.

The condition $\alpha \vec{a} + \beta \vec{b} + \gamma \vec{c} = \vec{0} \Rightarrow \alpha = \beta = \gamma = 0$ implies that the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are linearly independent. For three vectors to be linearly independent in three-dimensional space, they must be non-coplanar, which is equivalent to the requirement that the determinant of the matrix formed by their components is non-zero. By arranging the given vector components as rows in a matrix, we obtain the following determinant expression:

$\begin{vmatrix} 1+t & 1-t & 1 \\ 1-t & 1+t & 2 \\ t & -t & 1 \end{vmatrix} \neq 0$

Expanding this determinant along the first row yields $(1+t)(1+t+2t) - (1-t)(1-t-2t) + 1((1-t)(-t) - (1+t)(t))$. Simplifying the algebraic expressions within this calculation leads to $(3t^{2+}4t+1) - (3t^{2-}4t+1) - 2t$, which further reduces to $6t$. Since this value must not be equal to zero, it follows that $t$ can be any real number excluding zero.

The vector identity $2(\vec{a} \times \vec{b}) = 3(\vec{c} \times \vec{a})$ implies that the vector $\vec{a}$ is parallel to $2\vec{b} + 3\vec{c}$, which allows the relationship $\vec{a} = \lambda(2\vec{b} + 3\vec{c})$ to be established. Squaring the magnitude yields $|\vec{a}|^2 = \lambda^2 (4|\vec{b}|^2 + 9|\vec{c}|^2 + 12\vec{b} \cdot \vec{c})$, and using the given magnitudes $|\vec{a}| = \sqrt{31}$, $|\vec{b}| = 0.5$, $|\vec{c}| = 2$, along with the dot product $\vec{b} \cdot \vec{c} = (0.5)(2) \cos(2\pi/3) = -0.5$, we find $31 = \lambda^2(4(0.25) + 9(4) + 12(-0.5))$, confirming $\lambda^2 = 1$.

Substituting $\vec{a} = \pm(2\vec{b} + 3\vec{c})$ into the target expression $\left(\frac{\vec{a}\times \vec{c}}{\vec{a}\cdot\vec{b}}\right)^2$ reduces the numerator to $\pm 2(\vec{b} \times \vec{c})$ and the denominator to $\pm(2|\vec{b}|^2 + 3\vec{b} \cdot \vec{c}) = \pm(2(0.25) + 3(-0.5)) = \mp 1$. Squaring this ratio results in $4|\vec{b} \times \vec{c}|^2$, and calculating the squared magnitude as $|\vec{b}|^2|\vec{c}|^2\sin^2(2\pi/3) = (0.25)(4)(0.75) = 0.75$ leads to a final value of $4 \times 0.75 = 3$.

Calculating the unknown vector $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ relies on solving a system of linear equations derived from the given vector constraints. First, computing the cross product $\vec{a} \times \vec{b} = 15\hat{i} - 20\hat{j} - 25\hat{k}$ allows us to simplify the equation $\vec{c} \cdot (\vec{a} \times \vec{b}) + 25 = 0$ to $3x - 4y - 5z = -5$. By applying the remaining conditions $x + y + z = 4$ and the projection constraint $\frac{\vec{c} \cdot \vec{a}}{|\vec{a}|} = 1$, which simplifies to $4x + 3y = 5$, we identify the coordinates of the vector as $\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$. Finally, the projection of $\vec{c}$ onto $\vec{b}$ is determined using the dot product formula $\frac{\vec{c} \cdot \vec{b}}{|\vec{b}|}$, which results in $\frac{25}{5\sqrt{2}} = \frac{5}{\sqrt{2}}$.

Resolving vector equations requires converting cross products into component-wise algebraic relations. Calculating the sum $\vec{a} + \vec{b} = 5\hat{i} + \hat{j} + 4\hat{k}$ and the cross product $\vec{a} \times \vec{b} = 7\hat{i} - 7\hat{j} - 7\hat{k}$ simplifies the primary vector equation. Substituting these into the original expression gives $(5\hat{i} + \hat{j} + 4\hat{k}) \times \vec{c} = 14\hat{i} + 10\hat{j} - 20\hat{k}$. By setting $\vec{c} = x\hat{i} + y\hat{j} + z\hat{k}$ and expanding the cross product using a determinant, the components yield the system $z - 4y = 14$, $4x - 5z = 10$, and $5y - x = -20$.

The scalar equation $(\vec{a} - \vec{b} + \hat{i}) \cdot \vec{c} = -3$ simplifies to $(2\hat{i} + 3\hat{j} - 2\hat{k}) \cdot \vec{c} = -3$, which provides the linear relation $2x + 3y - 2z = -3$. From the cross product relations, $x = 5y + 20$. Substituting this into the scalar equation results in $2(5y + 20) + 3y - 2z = -3$, which further simplifies to $13y - 2z = -43$. Using $z = 4y + 14$ from the first cross product component relation, the equation becomes $13y - 2(4y + 14) = -43$. Solving this leads to $5y = -15$, so $y = -3$. Consequently, $z = 2$ and $x = 5$. The squared magnitude $|\vec{c}|^2$ is then $5^2 + (-3)^2 + 2^2$, resulting in 38.

Placing the vectors on a Cartesian plane with the center $O$ at the origin simplifies the geometry, where we define $\vec{OP} = \vec{u} = \hat{i}$ and $\vec{OQ} = \hat{j}$ since the arc subtends a right angle. Given $R$ is the midpoint of the arc, $\vec{OR} = \vec{v}$ bisects the angle between the vectors, resulting in $\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$. Expressing $\vec{OQ}$ as the linear combination $\alpha\vec{u} + \beta\vec{v}$ leads to the equation $\hat{j} = \alpha\hat{i} + \beta(\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j})$. By equating the coefficients of $\hat{i}$ and $\hat{j}$, we obtain $\alpha + \frac{\beta}{\sqrt{2}} = 0$ and $\frac{\beta}{\sqrt{2}} = 1$, which yield $\beta = \sqrt{2}$ and $\alpha = -1$. These values provide roots $\alpha = -1$ and $\beta^2 = 2$ for the required quadratic equation, expressed as $x^2 - (\alpha + \beta^2)x + \alpha\beta^2 = 0$, which simplifies to $x^2 - x - 2 = 0$.

The coplanarity of three vectors requires their scalar triple product to be zero, meaning the determinant of the matrix formed by their components vanishes:
$\begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1 \end{vmatrix} = 0$
Expanding this determinant yields $10\lambda - 2\mu - 56 = 0$, which simplifies to the linear equation $5\lambda - \mu = 28$. The projection of $\vec{a}$ onto $\vec{b}$ is defined as the quotient of the dot product $\vec{a} \cdot \vec{b}$ and the magnitude $|\vec{b}|$. With $|\vec{b}| = \sqrt{24}$ and the scalar projection given as $\sqrt{54}$, the relationship is:
$\frac{-2\lambda + 4\mu - 8}{\sqrt{24}} = \sqrt{54}$
Multiplying by $\sqrt{24}$ yields $-2\lambda + 4\mu - 8 = 36$, which further simplifies to $-\lambda + 2\mu = 22$. Solving this system of linear equations for $\lambda$ and $\mu$ results in a total sum of $\lambda + \mu = 24$.

The vector triple product expansion allows us to rewrite the expression as $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Given that this expression equals $\vec{b} + \lambda\vec{c}$ and that the vectors $\vec{b}$ and $\vec{c}$ are non-parallel, we can determine the value of $\lambda$ by equating the coefficients of these linearly independent vectors on both sides of the equation. This comparison establishes the relations $\vec{a} \cdot \vec{c} = 1$ and $-\vec{a} \cdot \vec{b} = \lambda$. Calculating the dot product of $\vec{a} = 3\hat{i}+\hat{j}$ and $\vec{b} = \hat{i}+2\hat{j}+\hat{k}$ provides $(3)(1) + (1)(2) + (0)(1) = 5$. By substituting this result into the equality $-\vec{a} \cdot \vec{b} = \lambda$, we find that $-(5) = \lambda$, which identifies the value of $\lambda$ as $-5$.

The geometric relationship between two vectors is governed by their dot product, which is defined as the product of their magnitudes and the cosine of the angle between them. Since $\hat{u}$ is a unit vector and each of the given reference vectors also has a magnitude of unity, the projection of $\hat{u}$ onto each reference vector is simply the cosine of the specified angle, allowing the system to be expressed through a set of linear equations.

Let $\vec{p}_1 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$, $\vec{p}_2 = \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$, and $\vec{p}_3 = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j}$. Applying the dot product condition $\hat{u} \cdot \vec{p} = \cos(\theta)$ produces the following equations: $\hat{u} \cdot \vec{p}_1 = 0$ leads to $\frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0$, or $x+z=0$; $\hat{u} \cdot \vec{p}_2 = \frac{1}{2}$ leads to $\frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2}$, or $y+z=\frac{1}{\sqrt{2}}$; and $\hat{u} \cdot \vec{p}_3 = -\frac{1}{2}$ leads to $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2}$, or $x+y=-\frac{1}{\sqrt{2}}$. Substituting $z = -x$ into the second equation gives $y - x = \frac{1}{\sqrt{2}}$. Summing this with the third equation $x + y = -\frac{1}{\sqrt{2}}$ yields $2y = 0$, meaning $y=0$. Consequently, $z = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$, which determines the vector as $\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{k}$.

Given the vector $\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$, the difference vector is $\hat{u} - \vec{v} = (-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}})\hat{k}$, simplifying to $\hat{u} - \vec{v} = -\frac{2}{\sqrt{2}}\hat{i} - \frac{1}{\sqrt{2}}\hat{j}$. The square of the magnitude is calculated as $|\hat{u}-\vec{v}|^2 = (-\frac{2}{\sqrt{2}})^2 + (-\frac{1}{\sqrt{2}})^2 = \frac{4}{2} + \frac{1}{2} = \frac{5}{2}$.

The decomposition of a vector $\vec{\beta}$ into components parallel and perpendicular to a reference vector $\vec{\alpha}$ relies on the projection formula. Because $\vec{\beta}_1$ is parallel to $\vec{\alpha}$, it must take the form $\lambda \vec{\alpha}$ for some scalar $\lambda$, while the orthogonality condition $\vec{\beta}_2 \cdot \vec{\alpha} = 0$ allows us to uniquely determine $\lambda$ using the relation $\vec{\beta}_2 = \vec{\beta} - \lambda \vec{\alpha}$.

By setting $(\vec{\beta} - \lambda \vec{\alpha}) \cdot \vec{\alpha} = 0$, we derive the expression $\lambda = \frac{\vec{\beta} \cdot \vec{\alpha}}{|\vec{\alpha}|^2}$. Substituting the given vectors, the dot product $\vec{\beta} \cdot \vec{\alpha} = (1)(4) + (2)(3) + (-4)(5) = -10$ and the magnitude squared $|\vec{\alpha}|^2 = 4^2 + 3^2 + 5^2 = 50$ yield $\lambda = -\frac{1}{5}$. The perpendicular component is then $\vec{\beta}_2 = (\hat{i} + 2\hat{j} - 4\hat{k}) - (-\frac{1}{5})(4\hat{i} + 3\hat{j} + 5\hat{k})$, which simplifies to $\frac{9}{5}\hat{i} + \frac{13}{5}\hat{j} - 3\hat{k}$. Multiplying this result by 5 gives the vector $9\hat{i} + 13\hat{j} - 15\hat{k}$, and the final dot product with the sum vector $\hat{i} + \hat{j} + \hat{k}$ results in $9 + 13 - 15$, which equals 7.