Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

Squaring the magnitude equation $|\hat{a}+\hat{b}+2(\hat{a}\times \hat{b})|=2$ reveals the relationship between the vectors and the angle $\theta$ between them. Since $(\hat{a}+\hat{b})$ is orthogonal to $(\hat{a}\times \hat{b})$, the squared magnitude distributes as:
$|\hat{a}+\hat{b}|^2 + |2(\hat{a}\times \hat{b})|^2 = 4$
Expanding these terms yields $|\hat{a}|^2 + |\hat{b}|^2 + 2\hat{a}\cdot\hat{b} + 4|\hat{a}|^2|\hat{b}|^2\sin^2\theta = 4$. Substituting the unit vector magnitudes results in $2 + 2\cos\theta + 4\sin^2\theta = 4$. Replacing $\sin^2\theta$ with $1-\cos^2\theta$ transforms this into the quadratic expression $2\cos^2\theta - \cos\theta - 1 = 0$, which factors into $(2\cos\theta+1)(\cos\theta-1)=0$. Given the restriction $\theta \in (0,\pi)$, $\cos\theta$ must equal $-0.5$, which corresponds to $\theta = 2\pi/3$.

Comparing $2|\hat{a}\times \hat{b}|$ to $|\hat{a}-\hat{b}|$ confirms the validity of the first statement. With $\theta = 2\pi/3$, the magnitude of the cross product is $\sin(2\pi/3) = \sqrt{3}/2$, making $2|\hat{a}\times \hat{b}| = \sqrt{3}$. Simultaneously, the magnitude of the difference vector is $\sqrt{|\hat{a}|^2 + |\hat{b}|^2 - 2\hat{a}\cdot\hat{b}} = \sqrt{1 + 1 - 2(-0.5)} = \sqrt{3}$. As both calculations yield the same result, the first assertion is true.

The projection of $\hat{a}$ onto $(\hat{a}+\hat{b})$ is defined by the ratio $\frac{\hat{a} \cdot (\hat{a}+\hat{b})}{|\hat{a}+\hat{b}|}$. The dot product in the numerator simplifies to $|\hat{a}|^2 + \hat{a}\cdot\hat{b} = 1 - 0.5 = 0.5$. The denominator magnitude is $\sqrt{|\hat{a}|^2 + |\hat{b}|^2 + 2\hat{a}\cdot\hat{b}} = \sqrt{1 + 1 - 1} = 1$. Consequently, the ratio is $0.5/1 = 0.5$, indicating that the second statement is also accurate.

The angle between two vectors is acute if and only if their dot product is strictly positive. Calculating the dot product of $\vec{u}$ and $\vec{v}$ results in the expression $a(\log_e b)^2 - 12 + 6a(\log_e b)$, which must be greater than zero. Given that $b > 1$, we can let $t = \log_e b$, where $t > 0$. This transforms the condition into the inequality $at^2 + 6at - 12 > 0$ that must hold for all $t > 0$.

Analyzing this quadratic expression in $t$, we can see that if $a \leq 0$, the function will be negative for large values of $t$, so no such $a$ can satisfy the requirement. If $a > 0$, the parabola opens upward with a vertex at $t = -3$. Since the vertex lies outside the required domain of $t > 0$, the function is strictly increasing for all positive $t$. Because the function takes the value $-12$ at $t = 0$ and increases from that point, it will necessarily produce negative values for sufficiently small positive $t$. Therefore, no real value of $a$ satisfies the condition for all $t > 0$, leaving the set $S$ as the empty set.

Representing the unknown vector as $\vec{a} = x \hat{i} + y \hat{j} + z \hat{k}$ allows for the direct evaluation of the cross product through a determinant, yielding $\vec{a} \times \vec{b} = (\lambda y - z) \hat{i} + (z - \lambda x) \hat{j} + (x - y) \hat{k}$. By equating this to the specified result $13 \hat{i} - \hat{j} - 4 \hat{k}$, we obtain the system of linear equations $\lambda y - z = 13$, $z - \lambda x = -1$, and $x - y = -4$. Combined with the dot product relation $\vec{a} \cdot \vec{b} + 21 = 0$, which translates to $x + y + \lambda z = -21$, these equations allow us to uniquely determine the scalar components and the value of $\lambda$.

Solving the system yields $\lambda = 3$, $x = -2$, $y = 2$, and $z = -7$, which fully defines the vectors as $\vec{a} = -2 \hat{i} + 2 \hat{j} - 7 \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + 3 \hat{k}$. Substituting these into the required expression, we calculate $\vec{b} - \vec{a} = 3 \hat{i} - \hat{j} + 10 \hat{k}$ and $\vec{b} + \vec{a} = -\hat{i} + 3 \hat{j} - 4 \hat{k}$. Computing the individual dot products results in $(\vec{b} - \vec{a}) \cdot (-\hat{j} + \hat{k}) = 1 + 10 = 11$ and $(\vec{b} + \vec{a}) \cdot (\hat{i} - \hat{k}) = -1 + 4 = 3$. Summing these values gives a final result of 14.

We start by identifying the magnitude of vector $\vec{a}$ using the dot product formula $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$. Given the angle $\theta = \frac{\pi}{4}$, $|\vec{b}|^2 = 6$, and $\vec{a} \cdot \vec{b} = 3\sqrt{2}$, we substitute these into the equation to obtain $|\vec{a}| \sqrt{6} \cos\left(\frac{\pi}{4}\right) = 3\sqrt{2}$. Since $\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}$, this simplifies to $|\vec{a}| \sqrt{6} \cdot \frac{1}{\sqrt{2}} = |\vec{a}| \sqrt{3} = 3\sqrt{2}$, which confirms that $|\vec{a}| = \sqrt{6}$. Squaring both sides yields $|\vec{a}|^2 = 6$, and since $\vec{a} = \hat{i} + \alpha \hat{j} + \beta \hat{k}$, we have $|\vec{a}|^2 = 1 + \alpha^2 + \beta^2$. Therefore, $1 + \alpha^2 + \beta^2 = 6$, which implies that $\alpha^2 + \beta^2 = 5$.

To determine the value of $(\alpha^2 + \beta^2) |\vec{a} \times \vec{b}|^2$, we recall that the magnitude of the cross product is defined as $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$. Squaring this relationship gives $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta$. Substituting our known values of $|\vec{a}|^2 = 6$, $|\vec{b}|^2 = 6$, and $\sin^2\left(\frac{\pi}{4}\right) = \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2}$, we compute $|\vec{a} \times \vec{b}|^2 = 6 \times 6 \times \frac{1}{2} = 18$. Multiplying this result by $\alpha^2 + \beta^2 = 5$ gives a final result of $5 \times 18 = 90$.

In any closed triangle, the vectors representing the sides satisfy the relation $\vec{a} + \vec{b} + \vec{c} = 0$, which allows us to write $\vec{a} + \vec{c} = -\vec{b}$. To evaluate the first statement, we simplify the expression $|\vec{a}\times \vec{b}+\vec{c}\times \vec{b}|$ by factoring the cross product as $|(\vec{a}+\vec{c})\times \vec{b}|$, which equals $|(-\vec{b})\times \vec{b}| = 0$. Given the vector magnitudes and the dot product $\vec{b} \cdot \vec{c} = 12$, we determine the magnitude of $\vec{c}$ using $|\vec{c}|^2 = |\vec{a}+\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = 72 + 12 + 2(-24) = 36$, so $|\vec{c}| = 6$. Since the magnitude of the cross product portion is zero, the left side of the first statement simplifies to $-6$, which does not equal the value $6(2\sqrt{2}-1)$, proving the first statement is false.

For the second statement, the angle $\angle ACB$ is the angle between vectors $\vec{CA} = \vec{b}$ and $\vec{CB} = -\vec{a}$. Using the identity $\vec{a} + \vec{b} + \vec{c} = 0$, we find that $\vec{c} = -(\vec{a} + \vec{b})$, and substituting this into the dot product $\vec{b} \cdot \vec{c} = 12$ gives $\vec{b} \cdot (-\vec{a} - \vec{b}) = 12$, which simplifies to $-\vec{b} \cdot \vec{a} - 12 = 12$, resulting in $\vec{b} \cdot \vec{a} = -24$. The cosine of the angle $\angle ACB$ is calculated as $\frac{\vec{CA} \cdot \vec{CB}}{|\vec{CA}||\vec{CB}|} = \frac{\vec{b} \cdot (-\vec{a})}{|\vec{b}||\vec{a}|} = \frac{-\vec{b} \cdot \vec{a}}{|\vec{b}||\vec{a}|} = \frac{24}{(2\sqrt{3})(6\sqrt{2})} = \frac{24}{12\sqrt{6}} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}$. This confirms that the second statement correctly identifies the angle.

The core principle here relies on the geometric property that the cross product of two vectors produces a vector orthogonal to both inputs. Because the term $\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}$ is perpendicular to $\overset{\rightarrow}{b}$, the dot product of these two vectors is necessarily zero.

To find the value of $\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{c}$, take the dot product of $\overset{\rightarrow}{b}$ with the given expression for $\overset{\rightarrow}{c}$:
$\overset{\rightarrow}{b} \cdot \overset{\rightarrow}{c} = \overset{\rightarrow}{b} \cdot (2 \overset{\rightarrow}{a} \times \overset{\rightarrow}{b}) - 3 \overset{\rightarrow}{b} \cdot \overset{\rightarrow}{b}$
Since the first term involves a scalar triple product where the vector $\overset{\rightarrow}{b}$ is crossed with $\overset{\rightarrow}{a}$ and then dotted with itself, that portion of the expression simplifies to zero. This leaves us with the calculation $-3 \left| \overset{\rightarrow}{b} \right|^2$. Substituting the known magnitude of $\overset{\rightarrow}{b}$ as 4, we evaluate $-3 \times 16$, which results in -48.

Vector cross products distribute over addition, which allows the expression $(\vec{a}+2\vec{b}) \times (2\vec{a}-3\vec{b})$ to be expanded by multiplying each term while respecting the order of the vectors. Since the cross product of any vector with itself is zero, the expression simplifies to $-3(\vec{a} \times \vec{b}) + 4(\vec{b} \times \vec{a})$. Recalling that $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$, this simplifies further to:
$-3(\vec{a} \times \vec{b}) - 4(\vec{a} \times \vec{b}) = -7(\vec{a} \times \vec{b})$
Taking the squared magnitude of this result gives $49 |\vec{a} \times \vec{b}|^2$. By the geometric definition $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta$, we calculate the magnitude using the given values $|\vec{a}|=2$, $|\vec{b}|=3$, and $\theta = \frac{\pi}{4}$ to get $2 \times 3 \times \frac{1}{\sqrt{2}} = 3\sqrt{2}$. Squaring this value yields $18$, and multiplying $18$ by $49$ provides the final result of $882$.

Three vectors are coplanar if and only if their scalar triple product is zero, a condition which is equivalent to stating that the determinant of the coefficients of their basis vectors $\vec{a}, \vec{b}$, and $\vec{c}$ is zero. Given the position vectors $A = \vec{a}-\vec{b}+\vec{c}$, $B = \lambda \vec{a}-3 \vec{b}+4 \vec{c}$, $C = -\vec{a}+2 \vec{b}-3 \vec{c}$, and $D = 2 \vec{a}-4 \vec{b}+6 \vec{c}$, the displacement vectors are $\vec{AB} = (\lambda-1)\vec{a}-2\vec{b}+3\vec{c}$, $\vec{AC} = -2\vec{a}+3\vec{b}-4\vec{c}$, and $\vec{AD} = \vec{a}-3\vec{b}+5\vec{c}$.

Setting the determinant of these components equal to zero leads to the equation:
$\begin{vmatrix} \lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix}=0$
Expanding this determinant along the first row gives $(\lambda-1)(15-12) + 2(-10+4) + 3(6-3)=0$, which simplifies to $3(\lambda-1) - 12 + 9 = 0$. Solving this results in $3\lambda - 6 = 0$, yielding $\lambda = 2$.

Points $A, B,$ and $C$ are collinear if the vector $\vec{AB}$ is parallel to $\vec{AC}$, which mathematically means their corresponding direction ratios must be identical. Subtracting the position vectors, we find $\vec{AB} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$ and $\vec{AC} = 2\hat{i} - 6\hat{j} + 2\hat{k}$. Setting the ratios equal, $\frac{1}{2} = \frac{\alpha-4}{-6} = \frac{1}{2}$, reveals that the points are collinear specifically when $\alpha = 1$. Because the problem states the points must be non-collinear, we select the smallest positive integer for $\alpha$ that avoids collinearity, which is $\alpha = 2$.

With $\alpha = 2$, point $B$ is represented by the position vector $2\hat{i} + 2\hat{j} + 4\hat{k}$. The median from $A$ to the opposite side $BC$ bisects segment $BC$ at midpoint $M$, determined by averaging the coordinates of $B$ and $C$ to get $M = \left(\frac{5}{2}, 0, \frac{9}{2}\right)$. The length of the median $AM$ is the distance between $A(1, 4, 3)$ and $M\left(\frac{5}{2}, 0, \frac{9}{2}\right)$, calculated as $\sqrt{(\frac{5}{2} - 1)^2 + (0 - 4)^2 + (\frac{9}{2} - 3)^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{18}{4} + \frac{64}{4}} = \frac{\sqrt{82}}{2}$.

The scalar projection of a vector $\vec{u}$ onto a non-zero vector $\vec{v}$ is defined by the formula $\frac{\vec{u} \cdot \vec{v}}{|\vec{v}|}$. To begin, the unknown scalars $\alpha$ and $\beta$ within vector $\vec{a}$ are determined by evaluating the given cross product

. Expanding this determinant yields $(4 + 5\beta)\hat{i} - (4\alpha - 3\beta)\hat{j} + (-5\alpha - 3)\hat{k}$, which must equal $-\hat{i} + 9\hat{j} + 12\hat{k}$. By equating the corresponding components, we find $4 + 5\beta = -1$, yielding $\beta = -1$, and $-5\alpha - 3 = 12$, yielding $\alpha = -3$.

With these values, the vectors are defined as $\vec{a} = -3\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 3\hat{i} - 5\hat{j} + 4\hat{k}$. The two vectors required for the projection are $\vec{u} = \vec{b} - 2\vec{a} = 9\hat{i} - 7\hat{j} + 6\hat{k}$ and $\vec{v} = \vec{b} + \vec{a} = -4\hat{j} + 3\hat{k}$. Calculating the dot product $\vec{u} \cdot \vec{v}$ gives $(9)(0) + (-7)(-4) + (6)(3) = 28 + 18 = 46$, and the magnitude of the target vector is $|\vec{v}| = \sqrt{0^2 + (-4)^2 + 3^2} = 5$. Dividing the dot product by the magnitude results in the final projection value of $\frac{46}{5}$.

The area of a parallelogram defined by diagonal vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$ is expressed as $\frac{1}{2} |\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}| = 2\sqrt{2}$, which implies the magnitude of their cross product is $4\sqrt{2}$. Given that the magnitudes of the dot product and cross product are equal, we can write $|\overset{\rightarrow}{a}||\overset{\rightarrow}{b}|\cos \theta = |\overset{\rightarrow}{a}||\overset{\rightarrow}{b}|\sin \theta$, which simplifies to $\tan \theta = 1$. Since the angle $\theta$ is acute, it must be $\frac{\pi}{4}$, and substituting this into $|\overset{\rightarrow}{a}||\overset{\rightarrow}{b}|\sin \theta = 4\sqrt{2}$ with $|\overset{\rightarrow}{a}|=1$ yields $|\overset{\rightarrow}{b}| = 8$.

To find the angle $\alpha$ between $\overset{\rightarrow}{b}$ and $\overset{\rightarrow}{c}$, we compute their dot product and magnitudes. Taking the dot product of $\overset{\rightarrow}{c} = 2\sqrt{2}(\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}) - 2\overset{\rightarrow}{b}$ with $\overset{\rightarrow}{b}$ eliminates the cross product term because it is orthogonal to $\overset{\rightarrow}{b}$, resulting in $\overset{\rightarrow}{c} \cdot \overset{\rightarrow}{b} = -2|\overset{\rightarrow}{b}|^2 = -128$. The squared magnitude of $\overset{\rightarrow}{c}$ is calculated as $|\overset{\rightarrow}{c}|^2 = 8|\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}|^2 + 4|\overset{\rightarrow}{b}|^2$, which equals $8(32) + 4(64) = 512$, so $|\overset{\rightarrow}{c}| = 16\sqrt{2}$. The cosine of the angle is then $\cos \alpha = \frac{\overset{\rightarrow}{c} \cdot \overset{\rightarrow}{b}}{|\overset{\rightarrow}{c}| |\overset{\rightarrow}{b}|} = \frac{-128}{(16\sqrt{2})(8)} = -\frac{1}{\sqrt{2}}$, leading to $\alpha = \frac{3\pi}{4}$.

Expanding the cross product term $(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})$ simplifies efficiently because the cross product of any vector with itself is zero, and $\vec{b} \times \vec{a}$ is the negative of $\vec{a} \times \vec{b}$. This reduces the expression to $2(\vec{a} \times \vec{b})$, and squaring its magnitude results in $4|\vec{a} \times \vec{b}|^2$. Combining this with the remaining part of the original expression gives $4|\vec{a} \times \vec{b}|^2 + 4(\vec{a} \cdot \vec{b})^2$, which factors into $4(|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2)$. By applying Lagrange's Identity, the sum inside the parentheses is equivalent to $|\vec{a}|^2 |\vec{b}|^2$. Substituting the given magnitudes $|\vec{a}|=4$ and $|\vec{b}|=3$ into the simplified expression $4|\vec{a}|^2 |\vec{b}|^2$ leads to $4 \times 16 \times 9$, which equals 576.

A vector that is parallel to the line of intersection of two planes must be perpendicular to the normal vectors of both planes. By computing the cross product of the vectors spanning each plane, we identify the normal vectors $\vec{n}_1 = (\hat{i}+\hat{j}) \times (\hat{i}+\hat{k}) = \hat{i}-\hat{j}-\hat{k}$ and $\vec{n}_2 = (\hat{i}-\hat{j}) \times (\hat{j}-\hat{k}) = \hat{i}+\hat{j}+\hat{k}$. Since the vector $\vec{a}$ is parallel to the intersection of the planes, it is collinear with $\vec{n}_1 \times \vec{n}_2$. Calculating this cross product yields $-2\hat{j} + 2\hat{k}$, allowing us to express the vector as $\vec{a} = \lambda(-2\hat{j} + 2\hat{k})$.

Applying the given condition $\vec{a} \cdot \vec{b} = 6$ with $\vec{b} = 2\hat{i}-2\hat{j}+\hat{k}$, we compute the dot product $\lambda(-2\hat{j} + 2\hat{k}) \cdot (2\hat{i}-2\hat{j}+\hat{k}) = \lambda(0 + 4 + 2) = 6\lambda$, which results in $6\lambda = 6$ and thus $\lambda = 1$. This identifies the vector $\vec{a} = -2\hat{j} + 2\hat{k}$. To find the angle $\theta$ between $\vec{a}$ and $\vec{b}$, we use the relation $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$. With $|\vec{a}| = \sqrt{(-2)^2 + 2^2} = 2\sqrt{2}$ and $|\vec{b}| = \sqrt{2^2 + (-2)^2 + 1^2} = 3$, we find $\cos \theta = \frac{6}{2\sqrt{2} \cdot 3} = \frac{1}{\sqrt{2}}$, which corresponds to $\theta = \frac{\pi}{4}$.

Finally, the magnitude of the cross product $|\vec{a} \times \vec{b}|$ is determined using the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$. Substituting the known values, we have $|\vec{a} \times \vec{b}|^2 = (2\sqrt{2})^2(3)^2 - 6^2 = 8 \cdot 9 - 36 = 72 - 36 = 36$, so $|\vec{a} \times \vec{b}| = 6$. The resulting ordered pair is $\left(\frac{\pi}{4}, 6\right)$.

The angle between two vectors $\vec{b}$ and $\vec{c}$ is determined by the cosine formula $\cos \theta = \frac{\vec{b} \cdot \vec{c}}{|\vec{b}| |\vec{c}|}$, which requires calculating their dot product and their respective magnitudes. Given the definition $\vec{c} = 2(\vec{a} \times \vec{b}) - 3\vec{b}$, we can isolate these components by using the properties of scalar products and vector orthogonality.

To find the dot product $\vec{b} \cdot \vec{c}$, we distribute $\vec{b}$ across the expression to get $2\vec{b} \cdot (\vec{a} \times \vec{b}) - 3\vec{b} \cdot \vec{b}$. Since the cross product $\vec{a} \times \vec{b}$ is inherently perpendicular to $\vec{b}$, the term $\vec{b} \cdot (\vec{a} \times \vec{b})$ vanishes, leaving us with $-3|\vec{b}|^2$. Substituting $|\vec{b}| = 4$, we find the value of the dot product is $-3(16) = -48$.

To determine the magnitude $|\vec{c}|$, we compute $|\vec{c}|^2 = \vec{c} \cdot \vec{c} = 4|\vec{a} \times \vec{b}|^2 + 9|\vec{b}|^2$. We calculate the squared magnitude of the cross product using the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2|\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$, which yields $1(16) - 4 = 12$. Substituting this into our expression for $|\vec{c}|^2$ results in $4(12) + 9(16) = 48 + 144 = 192$, meaning $|\vec{c}| = \sqrt{192} = 8\sqrt{3}$.

Substituting these derived values into the cosine formula gives $\cos \theta = \frac{-48}{4 \cdot 8\sqrt{3}}$, which simplifies to $\frac{-48}{32\sqrt{3}}$. Reducing this fraction leads to $-\frac{\sqrt{3}}{2}$, indicating that the angle is $\cos^{-1} \left( -\frac{\sqrt{3}}{2} \right)$.