Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

To decompose a vector $\vec{\beta}$ into two components relative to a reference vector $\vec{\alpha}$, we identify one component, $\vec{\beta_1}$, as parallel to $\vec{\alpha}$, meaning $\vec{\beta_1} = k\vec{\alpha}$ for some scalar $k$. The second component, $\vec{\beta_2}$, is perpendicular to $\vec{\alpha}$, satisfying the condition $\vec{\beta_2} \cdot \vec{\alpha} = 0$. Using the given relationship $\vec{\beta} = \vec{\beta_1} - \vec{\beta_2}$, we rearrange the expression to isolate the perpendicular component as $\vec{\beta_2} = \vec{\beta_1} - \vec{\beta}$.

Substituting $\vec{\beta_1} = k\vec{\alpha}$ into the orthogonality condition $(\vec{\beta_1} - \vec{\beta}) \cdot \vec{\alpha} = 0$, we establish that $k(\vec{\alpha} \cdot \vec{\alpha}) = \vec{\beta} \cdot \vec{\alpha}$. Given $\vec{\alpha} = 3\hat{i} + \hat{j}$ and $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}$, the calculation becomes $k(3^2 + 1^2) = (3)(2) + (1)(-1) + (0)(3)$, which simplifies to $10k = 5$, yielding $k = \frac{1}{2}$. Consequently, the parallel component is $\vec{\beta_1} = \frac{1}{2}(3\hat{i} + \hat{j}) = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}$, and the perpendicular component is $\vec{\beta_2} = (\frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}) - (2\hat{i} - \hat{j} + 3\hat{k}) = -\frac{1}{2}\hat{i} + \frac{3}{2}\hat{j} - 3\hat{k}$.

Finally, the cross product $\vec{\beta_1} \times \vec{\beta_2}$ is calculated using the determinant of the basis vectors and the components of $\vec{\beta_1}$ and $\vec{\beta_2}$. Performing this evaluation results in $\hat{i}(-\frac{3}{2} - 0) - \hat{j}(-\frac{9}{2} - 0) + \hat{k}(\frac{9}{4} + \frac{1}{4})$, which simplifies to $-\frac{3}{2}\hat{i} + \frac{9}{2}\hat{j} + \frac{5}{2}\hat{k}$. Factoring out the common constant, the final result is $\frac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k})$.

r_\vec{g} = $a/2\hat{i} + a/2 \hat{k}$ r_\vec{H} = $a/2 \hat{j}+a/2 \hat{k}$ r_\vec{H}-r_\vec{g} = $1/2 a(\hat{j}-\hat{i})$

The magnitude of the cross product is derived from the components of the resultant vector, calculated using the determinant, which yields the expression $r = \sqrt{(2+x)^{2} + (3-x)^{2} + 25}$. Expanding and simplifying these terms results in the quadratic function $r = \sqrt{2x^{2} - 2x + 38}$. Identifying the range of $r$ involves locating the minimum value of this quadratic expression by completing the square, which reveals that the expression $2x^{2} - 2x + 38$ reaches a minimum value of $37.5$. Since the magnitude $r$ is constrained by this minimum, it follows that $r \ge \sqrt{37.5}$, which simplifies precisely to $r \ge 5 \sqrt{\frac{3}{2}}$.

When three vectors lie in the same plane, their scalar triple product must be zero, meaning the determinant of the matrix formed by their components is equal to zero. Constructing this determinant from the coefficients of $\vec{a}$, $\vec{b}$, and $\vec{c}$ results in the cubic equation $\lambda^3 - 2\lambda^2 - 9\lambda + 18 = 0$. Factoring this expression as $(\lambda^2 - 9)(\lambda - 2) = 0$ identifies the potential values for $\lambda$ as $3$, $-3$, and $2$.

Since the problem requires a non-zero cross product $\vec{a} \times \vec{c}$, we must discard the cases where $\lambda$ is $3$ or $-3$, because those values make $\vec{a}$ and $\vec{c}$ parallel, which would force the cross product to be zero. Using $\lambda = 2$ provides the specific vectors $\vec{a} = \hat{i} + 2\hat{j} + 4\hat{k}$ and $\vec{c} = 2\hat{i} + 4\hat{j} + 3\hat{k}$. Evaluating the cross product as the determinant

yields the final result $-10\hat{i} + 5\hat{j}$.

A vector perpendicular to two others is found using the cross product, as the resulting vector is inherently orthogonal to the plane containing the initial vectors. Calculating the two vectors required gives $\vec{a}+\vec{b} = 4\hat{i}+4\hat{j}+0\hat{k}$ and $\vec{a}-\vec{b} = 2\hat{i}+0\hat{j}+4\hat{k}$. The cross product of these two vectors is determined by the determinant

, which evaluates to $16\hat{i}-16\hat{j}-8\hat{k}$. The magnitude of this vector is $\sqrt{16^{2+}(-16)^{2+}(-8)^2} = \sqrt{256+256+64} = 24$. Because the problem specifies a final magnitude of 12, we scale the cross product by $\frac{12}{24}$ or $\frac{1}{2}$, resulting in $8\hat{i}-8\hat{j}-4\hat{k}$, which simplifies to $4(2\hat{i}-2\hat{j}-\hat{k})$.

The volume of a parallelepiped defined by three vectors is determined by the absolute value of the scalar triple product, represented by the determinant of the matrix formed by the vector components. Organizing the coefficients of the given vectors $\hat{i}+\lambda\hat{j}+\hat{k}$, $\hat{j}+\lambda\hat{k}$, and $\lambda\hat{i}+\hat{k}$ into a $3 \times 3$ matrix and evaluating the determinant results in the polynomial function $V(\lambda) = \lambda^3 - \lambda + 1$.

Taking the derivative of this volume function produces $V'(\lambda) = 3\lambda^2 - 1$. The critical points occur where the derivative is zero, which corresponds to $\lambda = \pm \frac{1}{\sqrt{3}}$. Utilizing the second derivative test, $V''(\lambda) = 6\lambda$, shows that at $\lambda = \frac{1}{\sqrt{3}}$, the result is positive, confirming that this point marks the minimum volume.

The shortest distance from a point to a line is the length of the perpendicular segment from the point to the line, which forms the altitude of a right triangle created by the vector connecting a point on the line to the external point and the line's own direction vector. Starting with the point on the line at $A(2,3,-4)$ and the given point $P(-1,2,6)$, we define the vector $\vec{AP} = (3,1,-10)$ with a squared magnitude of $110$. To find the length of the base of this triangle, we calculate the scalar projection of $\vec{AP}$ onto the line's direction vector $\vec{v} = 6\hat{i}+3\hat{j}-4\hat{k}$, which has a magnitude $|\vec{v}| = \sqrt{6^{2+}3^{2+}(-4)^2} = \sqrt{61}$ and results in a projection length $AQ = \frac{|\vec{AP} \cdot \vec{v}|}{|\vec{v}|} = \frac{61}{\sqrt{61}} = \sqrt{61}$. Finally, by the Pythagorean theorem, the perpendicular distance is $PQ = \sqrt{AP^2 - AQ^2} = \sqrt{110 - 61} = \sqrt{49}$, which simplifies to 7.

According to the Angle Bisector Theorem, the internal angle bisector of $\angle A$ in $\triangle ABC$ divides the opposite side $BC$ in the ratio of the lengths of the adjacent sides $AB$ and $AC$. The position vector $\vec{d}$ of the point $D$ on $BC$ is given by the section formula $\vec{d} = \frac{AB \vec{c} + AC \vec{b}}{AB + AC}$.

First, calculate the lengths of the adjacent sides: $AB = |\vec{b} - \vec{a}| = |(2-4)\hat{i} + (3-7)\hat{j} + (4-8)\hat{k}| = \sqrt{(-2)^2 + (-4)^2 + (-4)^2} = \sqrt{36} = 6$ and $AC = |\vec{c} - \vec{a}| = |(2-4)\hat{i} + (5-7)\hat{j} + (7-8)\hat{k}| = \sqrt{(-2)^2 + (-2)^2 + (-1)^2} = \sqrt{9} = 3$.

Substituting these lengths and the given position vectors into the section formula, where the ratio $AB:AC = 6:3 = 2:1$, yields $\vec{d} = \frac{2\vec{c} + 1\vec{b}}{2+1} = \frac{2(2\hat{i}+5\hat{j}+7\hat{k}) + (2\hat{i}+3\hat{j}+4\hat{k})}{3} = \frac{1}{3}(6\hat{i} + 13\hat{j} + 18\hat{k})$.

The squared magnitude of the vector equation $\vec{a} + 2\vec{b} + 2\vec{c} = \vec{0}$ can be utilized by isolating the term involving $\vec{b}$ to obtain $2\vec{b} = -(\vec{a} + 2\vec{c})$. Squaring both sides of this equation results in $4|\vec{b}|^2 = |\vec{a} + 2\vec{c}|^2 = |\vec{a}|^2 + 4|\vec{c}|^2 + 4(\vec{a} \cdot \vec{c})$. Substituting the unit vector magnitudes $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$ gives $4(1) = 1 + 4(1) + 4(\vec{a} \cdot \vec{c})$, which simplifies to $4 = 5 + 4(\vec{a} \cdot \vec{c})$, yielding $\vec{a} \cdot \vec{c} = -1/4$.

Since $\vec{a} \cdot \vec{c} = |\vec{a}||\vec{c}|\cos\theta = \cos\theta$, we have $\cos\theta = -1/4$. The magnitude of the cross product is given by $|\vec{a} \times \vec{c}| = |\vec{a}||\vec{c}|\sin\theta = \sin\theta$. Using the identity $\sin\theta = \sqrt{1 - \cos^2\theta}$, the value becomes $\sqrt{1 - (-1/4)^2} = \sqrt{1 - 1/16} = \sqrt{15/16} = \frac{\sqrt{15}}{4}$.

Given $\vec{A} = \hat{i}+\hat{j}$ and $\vec{B} = 2\hat{i}-\hat{j}$, we calculate $\vec{A} \cdot \vec{B} = (1)(2) + (1)(-1) = 1$. Since $\vec{C}$ is coplanar, let $\vec{C} = x\vec{A} + y\vec{B}$.
The conditions $\vec{A} \cdot \vec{C} = 1$ and $\vec{B} \cdot \vec{C} = 1$ yield $x|\vec{A}|^2 + y(\vec{A} \cdot \vec{B}) = 1$ and $x(\vec{A} \cdot \vec{B}) + y|\vec{B}|^2 = 1$.
Substituting $|\vec{A}|^2 = 2$ and $|\vec{B}|^2 = 5$ gives the system: $2x + y = 1$ and $x + 5y = 1$.
Solving this system, we find $y = \frac{1}{9}$ and $x = \frac{4}{9}$.
Thus, $\vec{C} = \frac{4}{9}(\hat{i}+\hat{j}) + \frac{1}{9}(2\hat{i}-\hat{j}) = \frac{2}{3}\hat{i} + \frac{1}{3}\hat{j}$.
Finally, the magnitude is $|\vec{C}| = \sqrt{(\frac{2}{3})^2 + (\frac{1}{3})^2} = \sqrt{\frac{4}{9} + \frac{1}{9}} = \sqrt{\frac{5}{9}}$.

The vector triple product identity $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$ facilitates the determination of $\vec{b}$ by relating the given dot and cross products. Applying this identity to $\vec{a} \times \vec{b} = \vec{c}$ yields $\vec{a} \times \vec{c} = (\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2 \vec{b}$. Given $|\vec{a}|^2 = 1^2 + 1^2 + 1^2 = 3$ and $\vec{a} \cdot \vec{b} = 3$, the equation simplifies to $3\vec{b} = 3\vec{a} - (\vec{a} \times \vec{c})$.

Calculating the cross product

, we substitute these values into the rearranged equation to find $3\vec{b} = 3(\hat{i} + \hat{j} + \hat{k}) - (-2\hat{i} + \hat{j} + \hat{k}) = 5\hat{i} + 2\hat{j} + 2\hat{k}$. Finally, the magnitude is $|\vec{b}| = \frac{1}{3} \sqrt{5^2 + 2^2 + 2^2} = \frac{1}{3} \sqrt{33} = \sqrt{\frac{33}{9}} = \sqrt{\frac{11}{3}}$.

Since $\vec{u}$ lies in the plane of $\vec{a}$ and $\vec{b}$ and is perpendicular to $\vec{a}$, it must be proportional to the vector triple product $\vec{a} \times (\vec{a} \times \vec{b})$. Applying the vector identity $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$, we calculate the necessary components: the dot product $\vec{a} \cdot \vec{b} = (2)(0) + (3)(1) + (-1)(1) = 2$ and the magnitude squared $|\vec{a}|^2 = 2^2 + 3^2 + (-1)^2 = 14$. Substituting these values allows us to express the vector as $\vec{u} = \lambda(2\vec{a} - 14\vec{b})$.

To determine the scalar $\lambda$, we utilize the condition $\vec{u} \cdot \vec{b} = 24$. Substituting our expression for $\vec{u}$ yields $\lambda(2\vec{a} \cdot \vec{b} - 14|\vec{b}|^2) = 24$. With $\vec{b} = \hat{j} + \hat{k}$, we find $|\vec{b}|^2 = 0^2 + 1^2 + 1^2 = 2$, leading to the equation $\lambda(2(2) - 14(2)) = 24$, which simplifies to $\lambda(-24) = 24$ and results in $\lambda = -1$. Consequently, the vector is defined by $\vec{u} = 14\vec{b} - 2\vec{a}$, which expands to $14(\hat{j} + \hat{k}) - 2(2\hat{i} + 3\hat{j} - \hat{k}) = -4\hat{i} + 8\hat{j} + 16\hat{k}$. Computing the square of its magnitude gives $|\vec{u}|^2 = (-4)^2 + 8^2 + 16^2 = 16 + 64 + 256$, which equals 336.

The area of a parallelogram defined by its diagonal vectors $\vec{d_1}$ and $\vec{d_2}$ is calculated as half the magnitude of their cross product, using the formula $\frac{1}{2} |\vec{d_1} \times \vec{d_2}|$. For the given vectors $8\hat{i}-6\hat{j}$ and $3\hat{i}+4\hat{j}-12\hat{k}$, the cross product yields a resultant vector with components 72, 96, and 50. The magnitude of this vector is found by calculating $\sqrt{72^{2+}96^{2+}50^2}$, which simplifies to $\sqrt{5184+9216+2500}$, or $\sqrt{16900}$. This square root calculation results in a magnitude of 130, and dividing this value by 2 provides the final area of 65 square units.

In any triangle, the orthocentre, centroid, and circumcentre are collinear points that lie on the Euler line, with the centroid positioned such that it divides the segment connecting the orthocentre and circumcentre in a $2:1$ ratio. This geometric property allows us to express the position vector of the orthocentre, $\vec{h}$, using the centroid $\vec{g}$ and the circumcentre $\vec{p}$ through the vector equation $\vec{h} = 3\vec{g} - 2\vec{p}$. Since the centroid of a triangle with vertices $\vec{a}$, $\vec{b}$, and $\vec{c}$ is defined as $\vec{g} = \frac{\vec{a}+\vec{b}+\vec{c}}{3}$ and the circumcentre is given as $\vec{p} = \frac{\vec{a}+\vec{b}+\vec{c}}{4}$, we can substitute these directly into the relation. This leads to the expression $\vec{h} = 3\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right) - 2\left(\frac{\vec{a}+\vec{b}+\vec{c}}{4}\right)$, which simplifies to $(\vec{a}+\vec{b}+\vec{c}) - \frac{\vec{a}+\vec{b}+\vec{c}}{2}$, ultimately resulting in the position vector $\frac{\vec{a}+\vec{b}+\vec{c}}{2}$.

In a triangle where the vertex A forms a right angle, the vectors $\vec{AB}$ and $\vec{AC}$ must be perpendicular, meaning their dot product is zero. Calculating these vectors from the given position points, we find $\vec{AB} = -4\hat{i} + 2\hat{j} + (p+1)\hat{k}$ and $\vec{AC} = 2\hat{i} + (q-1)\hat{j} - 3\hat{k}$. Setting their dot product equal to zero yields the equation $(-4)(2) + (2)(q-1) + (p+1)(-3) = 0$. Expanding this expression results in $-8 + 2q - 2 - 3p - 3 = 0$, which simplifies to $2q - 3p - 13 = 0$. Rearranging this as $q = \frac{3}{2}p + \frac{13}{2}$ identifies the slope of the line as $\frac{3}{2}$. Because the slope is a positive value, the line makes an acute angle with the positive direction of the $x$-axis.

The scalar triple product of consecutive cross products satisfies a known geometric property where the scalar triple product $[\vec{a}\times \vec{b}\;\;\vec{b}\times \vec{c}\;\;\vec{c}\times \vec{a}]$ is identically equal to the square of the scalar triple product of the original vectors, written as $[\vec{a}\;\vec{b}\;\vec{c}]^2$. Evaluating the term $[\vec{b}\times \vec{c}]\times [\vec{c}\times \vec{a}]$ yields the vector $[\vec{a}\vec{b}\vec{c}] \vec{c}$, and taking the dot product of this result with $[\vec{a}\times \vec{b}]$ confirms that the expression simplifies to the square of the scalar triple product. By equating this established identity to the expression $\lambda [\vec{a}\;\vec{b}\;\vec{c}]^2$, the value of $\lambda$ is 1.

In any triangle, the median vector originating from a vertex to the midpoint of the opposite side is defined as the arithmetic mean of the two vectors representing the adjacent sides. Applying this principle to vertex A of triangle ABC, the median vector $\vec{AD}$ is given by the relation $\vec{AD} = \frac{1}{2}(\vec{AB} + \vec{AC})$, where D represents the midpoint of side BC.

Substituting the provided vectors into this equation gives $\vec{AD} = \frac{1}{2}((3\hat{i} + 4\hat{k}) + (5\hat{i} - 2\hat{j} + 4\hat{k}))$. Summing the corresponding components results in $\frac{1}{2}(8\hat{i} - 2\hat{j} + 8\hat{k})$, which simplifies the vector expression to $4\hat{i} - \hat{j} + 4\hat{k}$. The length of this median is determined by calculating the magnitude of the vector, which is $\sqrt{4^2 + (-1)^2 + 4^2}$. Simplifying this square root leads to $\sqrt{16 + 1 + 16}$, resulting in a final value of $\sqrt{33}$.

Two vectors are perpendicular if and only if their scalar dot product is equal to zero. Setting the product of the given vectors $\vec{c} = \vec{a} + 2\vec{b}$ and $\vec{d} = 5\vec{a} - 4\vec{b}$ to zero provides the necessary condition:

$(\vec{a} + 2\vec{b}) \cdot (5\vec{a} - 4\vec{b}) = 0$

Expanding this product results in $5(\vec{a} \cdot \vec{a}) - 4(\vec{a} \cdot \vec{b}) + 10(\vec{b} \cdot \vec{a}) - 8(\vec{b} \cdot \vec{b}) = 0$. Because $\vec{a}$ and $\vec{b}$ are unit vectors, their squared magnitudes are $1$, allowing the equation to simplify to $5 + 6(\vec{a} \cdot \vec{b}) - 8 = 0$. Solving this expression yields $\vec{a} \cdot \vec{b} = \frac{1}{2}$. Given that $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$, the result $\cos \theta = \frac{1}{2}$ confirms that the angle between the vectors is $\frac{\pi}{3}$.

The altitude from vertex $B$ to the side $AD$ is found by identifying the projection of point $B$ onto the line containing vector $\vec{AD}$, which defines the vector $\vec{AX}$. By calculating the projection of $\vec{q}$ onto $\vec{p}$, we obtain $\vec{AX} = \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p}$. Since the altitude vector $\vec{r}$ represents the directed segment $\vec{BX}$, we use the triangle law of addition to write $\vec{r} = \vec{BA} + \vec{AX}$. Given $\vec{BA} = -\vec{q}$, the altitude is expressed as $\vec{r} = -\vec{q} + \frac{\vec{p} \cdot \vec{q}}{\vec{p} \cdot \vec{p}} \vec{p}$.

Vector triple product identities provide the most direct path to isolating the vector $\vec{d}$ in this configuration. By taking the cross product of $\vec{a}$ with both sides of the given equation $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$, we transform the expression into a form where the triple product identity $\vec{a} \times (\vec{b} \times \vec{v}) = (\vec{a} \cdot \vec{v})\vec{b} - (\vec{a} \cdot \vec{b})\vec{v}$ can be applied to both sides.

Applying this identity, the equation expands to $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = (\vec{a} \cdot \vec{d})\vec{b} - (\vec{a} \cdot \vec{b})\vec{d}$. Since the problem defines $\vec{a} \cdot \vec{d} = 0$, the first term on the right side vanishes, simplifying the expression to $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = -(\vec{a} \cdot \vec{b})\vec{d}$. Isolating $\vec{d}$ involves rearranging these terms and dividing by $-(\vec{a} \cdot \vec{b})$, which yields the final result: