Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

The volume of a parallelopiped defined by three coterminus vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by the magnitude of their scalar triple product, which is equivalent to the determinant of the matrix formed by their components. Setting the determinant equal to $158$ leads to the equation:
$\begin{vmatrix} 1 & 1 & n \\ 2 & 4 & -n \\ 1 & n & 3 \end{vmatrix} = 158$
Expanding this matrix yields $1(12 + n^2) - 1(6 + n) + n(2n - 4) = 158$, which simplifies to the quadratic $3n^2 - 5n - 152 = 0$. Solving this equation for non-negative values results in $n=8$. Substituting this value back into the vectors, the dot product $\vec{b} \cdot \vec{c}$ using $\vec{b} = 2\hat{i} + 4\hat{j} - 8\hat{k}$ and $\vec{c} = \hat{i} + 8\hat{j} + 3\hat{k}$ is calculated as $2(1) + 4(8) + (-8)(3) = 2 + 32 - 24 = 10$.

When the cross product of two vectors is zero, as seen in the rearrangement $\vec{b} \times (\vec{c} - \vec{a}) = \vec{0}$, it implies that the vector difference $(\vec{c} - \vec{a})$ must be parallel to $\vec{b}$. This relationship allows us to write $(\vec{c} - \vec{a}) = \lambda \vec{b}$ for some scalar $\lambda$, meaning $\vec{c} = \vec{a} + \lambda \vec{b}$. Applying the condition $\vec{c} \cdot \vec{a} = 0$ to this expression yields $(\vec{a} + \lambda \vec{b}) \cdot \vec{a} = 0$, which simplifies to $|\vec{a}|^2 + \lambda (\vec{b} \cdot \vec{a}) = 0$.

Substituting the given vector components $\vec{a} = \hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$, we find $|\vec{a}|^2 = 1^2 + (-2)^2 + 1^2 = 6$ and $\vec{b} \cdot \vec{a} = (1)(1) + (-1)(-2) + (1)(1) = 4$. Inserting these values into the equation results in $6 + 4\lambda = 0$, which gives $\lambda = -\frac{3}{2}$. The dot product $\vec{c} \cdot \vec{b}$ is then determined by calculating $(\vec{a} + \lambda \vec{b}) \cdot \vec{b}$, which expands to $\vec{a} \cdot \vec{b} + \lambda |\vec{b}|^2$. With $|\vec{b}|^2 = 1^2 + (-1)^2 + 1^2 = 3$, the final result is $4 + (-\frac{3}{2})(3) = 4 - \frac{9}{2} = -\frac{1}{2}$.

Given $|\vec{P}+\vec{Q}| = |\vec{P}|$, squaring both sides yields $P^2 + Q^2 + 2PQ\cos\alpha = P^2$, which simplifies to $Q^2 + 2PQ\cos\alpha = 0$, implying $\cos\alpha = -\frac{Q}{2P}$.
Let $\vec{S} = 2\vec{P} + \vec{Q}$ be the new resultant. The angle $\theta$ that $\vec{S}$ makes with $\vec{Q}$ is given by $\tan\theta = \frac{(2P)\sin\alpha}{Q + (2P)\cos\alpha}$.
Substituting $\cos\alpha = -\frac{Q}{2P}$ into the denominator:
$\tan\theta = \frac{2P\sin\alpha}{Q + 2P\left(-\frac{Q}{2P}\right)} = \frac{2P\sin\alpha}{Q - Q} = \frac{2P\sin\alpha}{0}$
Since the denominator is zero, $\tan\theta \to \infty$, which implies $\theta = 90^\circ$.

The expansion of vector magnitude squared, specifically $|\vec{u} \pm \vec{v}|^2 = |\vec{u}|^2 + |\vec{v}|^2 \pm 2\vec{u} \cdot \vec{v}$, relates the squared difference of vectors to their dot product. Given that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, their squared magnitudes are all 1. Expanding the relationship:

$|\vec{a}-\vec{b}|^2 + |\vec{a}-\vec{c}|^2 = 8$

results in $|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} + |\vec{a}|^2 + |\vec{c}|^2 - 2\vec{a} \cdot \vec{c} = 8$. Substituting 1 for the squared magnitudes simplifies this to $1 + 1 - 2\vec{a} \cdot \vec{b} + 1 + 1 - 2\vec{a} \cdot \vec{c} = 8$, which reduces to $4 - 2(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}) = 8$. This confirms the dot product sum:

$\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c} = -2$

The target expression $|\vec{a}+2\vec{b}|^2 + |\vec{a}+2\vec{c}|^2$ expands to $|\vec{a}|^2 + 4|\vec{b}|^2 + 4\vec{a} \cdot \vec{b} + |\vec{a}|^2 + 4|\vec{c}|^2 + 4\vec{a} \cdot \vec{c}$. Grouping the terms yields $2|\vec{a}|^2 + 4|\vec{b}|^2 + 4|\vec{c}|^2 + 4(\vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c})$. Substituting the unit magnitudes of 1 and the previously calculated dot product sum of -2 into this expression gives $2(1) + 4(1) + 4(1) + 4(-2)$. This simplifies to $10 - 8$, which equals 2.00.

Point $P$ divides the line segment $AB$ internally in the ratio $\lambda:1$, so its position vector is given by $\vec{OP} = \frac{\lambda \vec{OB} + \vec{OA}}{\lambda+1}$. Substituting the given position vectors $\vec{OA} = \hat{i}+\hat{j}+\hat{k}$ and $\vec{OB} = 2\hat{i}+\hat{j}+3\hat{k}$, we find $\vec{OP} = \frac{\lambda(2\hat{i}+\hat{j}+3\hat{k}) + (\hat{i}+\hat{j}+\hat{k})}{\lambda+1} = \frac{(2\lambda+1)\hat{i} + (\lambda+1)\hat{j} + (3\lambda+1)\hat{k}}{\lambda+1}$.

Next, computing the dot product $\vec{OB}\cdot\vec{OP}$ yields $\frac{(2\hat{i}+\hat{j}+3\hat{k}) \cdot ((2\lambda+1)\hat{i} + (\lambda+1)\hat{j} + (3\lambda+1)\hat{k})}{\lambda+1} = \frac{2(2\lambda+1) + (\lambda+1) + 3(3\lambda+1)}{\lambda+1} = \frac{14\lambda+6}{\lambda+1}$. The cross product $\vec{OA}\times \vec{OP}$ is found using the determinant, where the vector components are $\hat{i}( \frac{3\lambda+1}{\lambda+1} - 1) - \hat{j}(\frac{3\lambda+1}{\lambda+1} - \frac{2\lambda+1}{\lambda+1}) + \hat{k}(1 - \frac{2\lambda+1}{\lambda+1}) = \frac{2\lambda}{\lambda+1}\hat{i} - \frac{\lambda}{\lambda+1}\hat{j} - \frac{\lambda}{\lambda+1}\hat{k}$. The squared magnitude of this cross product is $\frac{(2\lambda)^2 + (-\lambda)^2 + (-\lambda)^2}{(\lambda+1)^2} = \frac{6\lambda^2}{(\lambda+1)^2}$.

Substituting these results into the given equation $\vec{OB}\cdot\vec{OP}-3|\vec{OA}\times \vec{OP}|^{2}=6$ produces $\frac{14\lambda+6}{\lambda+1} - 3\left(\frac{6\lambda^2}{(\lambda+1)^2}\right) = 6$. Multiplying the entire equation by $(\lambda+1)^2$ leads to $(14\lambda+6)(\lambda+1) - 18\lambda^2 = 6(\lambda+1)^2$, which simplifies to $14\lambda^2 + 20\lambda + 6 - 18\lambda^2 = 6(\lambda^{2+}2\lambda+1)$. Further simplification gives $-4\lambda^2 + 20\lambda + 6 = 6\lambda^2 + 12\lambda + 6$, resulting in $10\lambda^2 - 8\lambda = 0$. Given that $\lambda > 0$, we find $\lambda = 0.8$.

Since vectors are coplanar if and only if their scalar triple product is zero, the determinant formed by their components must vanish. Evaluating the determinant

simplifies to $3a + 1 = 0$, which yields $a = -1/3$. Substituting this value, the dot product $\vec{p}\cdot\vec{q}$ calculates to $-1/3$, while the squared magnitudes are $|\vec{r}|^2 = 2/3$ and $|\vec{q}|^2 = 2/3$, with their dot product $\vec{r}\cdot\vec{q} = -1/3$. Applying Lagrange's identity $|\vec{r}\times \vec{q}|^2 = |\vec{r}|^2 |\vec{q}|^2 - (\vec{r}\cdot\vec{q})^2$, the squared cross product becomes $|\vec{r}\times \vec{q}|^2 = (2/3)(2/3) - (-1/3)^2 = 4/9 - 1/9 = 1/3$. Substituting these values into the given equation $3(\vec{p}\cdot\vec{q})^2 - \lambda |\vec{r}\times \vec{q}|^2 = 0$ results in $3(-1/3)^2 - \lambda(1/3) = 0$, which simplifies to $1/3 - \lambda/3 = 0$, leading to $\lambda = 1$.

Vector magnitude properties and the condition for orthogonality-specifically that two vectors are perpendicular if their dot product equals zero-determine the result. Squaring the given equation $\left|\overset{\rightarrow}{x}+\overset{\rightarrow}{y}\right|=\left|\overset{\rightarrow}{x}\right|$ leads to $\left|\overset{\rightarrow}{x}+\overset{\rightarrow}{y}\right|^2=\left|\overset{\rightarrow}{x}\right|^2$, which expands as follows:
$\left|\overset{\rightarrow}{x}\right|^{2+}\left|\overset{\rightarrow}{y}\right|^{2+}2\left(\overset{\rightarrow}{x}\cdot\overset{\rightarrow}{y}\right)=\left|\overset{\rightarrow}{x}\right|^2$
Subtracting $\left|\overset{\rightarrow}{x}\right|^2$ from both sides provides the essential relation $2\left(\overset{\rightarrow}{x}\cdot\overset{\rightarrow}{y}\right)=-\left|\overset{\rightarrow}{y}\right|^2$.

The condition that $2\overset{\rightarrow}{x}+\lambda\overset{\rightarrow}{y}$ is perpendicular to $\overset{\rightarrow}{y}$ implies their dot product $\left(2\overset{\rightarrow}{x}+\lambda\overset{\rightarrow}{y}\right)\cdot\overset{\rightarrow}{y}$ is zero. Distributing the dot product results in $2\left(\overset{\rightarrow}{x}\cdot\overset{\rightarrow}{y}\right)+\lambda\left|\overset{\rightarrow}{y}\right|^2=0$. Substituting the earlier relation $-\left|\overset{\rightarrow}{y}\right|^2$ for $2\left(\overset{\rightarrow}{x}\cdot\overset{\rightarrow}{y}\right)$ simplifies the expression to $-\left|\overset{\rightarrow}{y}\right|^{2+}\lambda\left|\overset{\rightarrow}{y}\right|^2=0$, which factors to $(\lambda-1)\left|\overset{\rightarrow}{y}\right|^2=0$. Because $\overset{\rightarrow}{y}$ is a non-zero vector, its magnitude squared is not zero, forcing $\lambda-1=0$, or $\lambda=1$.

The angle between two vectors $\vec{u}$ and $\vec{v}$ is defined by the relationship $\cos \phi = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$. Given the vectors $\vec{u} = a\hat{i} + b\hat{j} + c\hat{k}$ and $\vec{v} = b\hat{i} + c\hat{j} + a\hat{k}$, their magnitudes are identical, specifically $\sqrt{a^{2+}b^{2+}c^2} = 1$, which simplifies the cosine expression to the sum $ab+bc+ca$.

Setting the common value of the given equations equal to a constant $k$ allows us to express the variables as $a = \frac{k}{\cos \theta}$, $b = \frac{k}{\cos (\theta + 2\pi/3)}$, and $c = \frac{k}{\cos (\theta + 4\pi/3)}$. Substituting these values into the sum $ab+bc+ca$ leads to an expression proportional to the sum of the three cosine terms: $\cos \theta + \cos (\theta + 2\pi/3) + \cos (\theta + 4\pi/3)$. Because the sum of these specific cosine terms is zero for any angle $\theta$, the entire scalar product $ab+bc+ca$ vanishes, yielding $\cos \phi = 0$, which corresponds to an angle of $\frac{\pi}{2}$.

To isolate the unknown vector $\vec{c}$, we evaluate the equation $\vec{a} \times \vec{c} = -\vec{b}$ by taking the cross product of both sides with $\vec{a}$ from the right. This approach utilizes the vector triple product property, $(\vec{u} \times \vec{v}) \times \vec{w} = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{v} \cdot \vec{w})\vec{u}$. Applying this identity to the left side converts the expression into the following form:

$(\vec{a} \cdot \vec{a})\vec{c} - (\vec{c} \cdot \vec{a})\vec{a} = \vec{a} \times \vec{b}$

With $\vec{a} \cdot \vec{a} = 2$ and the given dot product $\vec{a} \cdot \vec{c} = 4$, the equation simplifies to $2\vec{c} - 4\vec{a} = \vec{a} \times \vec{b}$. Next, we calculate the cross product $\vec{a} \times \vec{b}$ using the determinant of the basis vectors, which yields $-\hat{i} - \hat{j} + 2\hat{k}$. Substituting $\vec{a} = \hat{i} - \hat{j}$ back into the equation results in $2\vec{c} = 4(\hat{i} - \hat{j}) + (-\hat{i} - \hat{j} + 2\hat{k})$. Simplifying the right side gives $2\vec{c} = 3\hat{i} - 5\hat{j} + 2\hat{k}$, which identifies the vector $\vec{c}$ as $\frac{3}{2}\hat{i} - \frac{5}{2}\hat{j} + \hat{k}$. Calculating the squared magnitude of this vector, we obtain:

$|\vec{c}|^2 = \left(\frac{3}{2}\right)^2 + \left(-\frac{5}{2}\right)^2 + 1^2 = \frac{9}{4} + \frac{25}{4} + 1 = \frac{38}{4} = \frac{19}{2}$

The projection of a vector $\vec{b}$ onto $\vec{a}$ is defined by the formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\vec{a}$. Given that this projection vector is $\vec{a}$ itself, the scalar component $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$ must be equivalent to the magnitude $|\vec{a}|$, which simplifies to the dot product condition $\vec{a} \cdot \vec{b} = |\vec{a}|^2$. Additionally, two vectors are perpendicular if and only if their dot product equals zero, a principle we use to constrain the sum of the vectors $\vec{a}+\vec{b}$ and $\vec{c}$.

Applying the projection condition, we calculate $|\vec{a}|^2 = 1^2 + 1^2 + (\sqrt{2})^2 = 4$ and the dot product $\vec{a} \cdot \vec{b} = b_1 + b_2 + 2$. Equating these values gives $b_1 + b_2 + 2 = 4$, which simplifies to $b_1 + b_2 = 2$. For the perpendicularity condition, we determine the vector $\vec{a} + \vec{b} = (1+b_1)\hat{i} + (1+b_2)\hat{j} + 2\sqrt{2}\hat{k}$ and compute its dot product with $\vec{c} = 5\hat{i} + \hat{j} + \sqrt{2}\hat{k}$ as $5(1+b_1) + 1(1+b_2) + \sqrt{2}(2\sqrt{2}) = 0$. This expands to $5 + 5b_1 + 1 + b_2 + 4 = 0$, resulting in the equation $5b_1 + b_2 = -10$. Solving the system formed by $b_1 + b_2 = 2$ and $5b_1 + b_2 = -10$ yields $4b_1 = -12$, so $b_1 = -3$ and $b_2 = 5$. The magnitude of $\vec{b}$ is then $\sqrt{(-3)^2 + 5^2 + (\sqrt{2})^2} = \sqrt{9 + 25 + 2} = \sqrt{36} = 6$.

The cosine of the angle between two vectors is given by their dot product divided by the product of their magnitudes. With $O$ as the origin, the centroid $G$ of the triangle with vertices $A(3, 0, -1)$, $B(2, 10, 6)$, and $C(1, 2, 1)$ can be found by calculating the average of the vertex coordinates, which yields $G(2, 4, 2)$.

The direction ratios for $\vec{OG}$ are $(1, 2, 1)$, and for $\vec{OA}$ are $(3, 0, -1)$. The calculation of the cosine proceeds as follows:
$\cos(\angle GOA) = \frac{(1)(3) + (2)(0) + (1)(-1)}{\sqrt{1^{2+}2^{2+}1^2} \sqrt{3^{2+}0^{2+}(-1)^2}}$

The numerator simplifies to $3 + 0 - 1 = 2$, and the product of the magnitudes in the denominator is $\sqrt{6} \times \sqrt{10} = \sqrt{60} = 2\sqrt{15}$. Dividing the numerator by the denominator gives $\frac{2}{2\sqrt{15}}$, which simplifies to $\frac{1}{\sqrt{15}}$.

Three vectors are coplanar if and only if their scalar triple product is zero, which is equivalent to saying the determinant of the matrix formed by their components vanishes. Applying this principle to the given vectors, we set the following determinant equal to zero:

Evaluating this determinant results in the equation $\mu(\mu^2 - 1) - 1(\mu - 1) + 1(1 - \mu) = 0$. Simplifying this expression systematically leads to the cubic equation $\mu^3 - 3\mu + 2 = 0$. Factoring the polynomial yields $(\mu - 1)^2(\mu + 2) = 0$, which identifies the roots $\mu = 1$ and $\mu = -2$. Summing these distinct real values gives a final result of -1.

Vectors are collinear if one can be expressed as a scalar multiple of the other, which implies that their components along linearly independent basis vectors must be proportional. For the vectors $\alpha^{\rightarrow} = (\lambda - 2)\vec{a} + \vec{b}$ and $\beta^{\rightarrow} = (4\lambda - 2)\vec{a} + 3\vec{b}$ to be collinear, the ratio of the coefficients of $\vec{a}$ must equal the ratio of the coefficients of $\vec{b}$, specifically $\frac{\lambda - 2}{4\lambda - 2} = \frac{1}{3}$. Cross-multiplying yields $3(\lambda - 2) = 4\lambda - 2$. Expanding this equation gives $3\lambda - 6 = 4\lambda - 2$. Solving for $\lambda$ by rearranging the terms results in $-4 = \lambda$.

The vector triple product expansion for $\vec{a} \times (\vec{b} \times \vec{c})$ is given by the identity $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. By setting this equal to $\frac{1}{2} \vec{b}$, we obtain the equation $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c} = \frac{1}{2} \vec{b} + 0 \vec{c}$. Given that $\vec{b}$ and $\vec{c}$ are non-parallel vectors, they are linearly independent, allowing us to equate the coefficients for each vector component individually.

This comparison directly shows that $\vec{a} \cdot \vec{c} = \frac{1}{2}$ and $\vec{a} \cdot \vec{b} = 0$. Recognizing that $\vec{a}$, $\vec{b}$, and $\vec{c}$ are unit vectors, the dot product definition $|\vec{a}| |\vec{b}| \cos \alpha = \vec{a} \cdot \vec{b}$ simplifies to $\cos \alpha = 0$, giving $\alpha = 90^\circ$. Similarly, $|\vec{a}| |\vec{c}| \cos \beta = \vec{a} \cdot \vec{c}$ simplifies to $\cos \beta = \frac{1}{2}$, which results in $\beta = 60^\circ$. Computing the difference between these two angles gives an absolute value of $|90^\circ - 60^\circ| = 30^\circ$.

$|\vec{A}+\vec{B}|$ = 2acosθ/2 ... (1) $|\vec{A}-\vec{B}|$ = 2acos $(\pi -\theta )/2$ = 2a sin θ/2 ... (2) \Rightarrow n $(2acos\theta /2)$ = 2a ${sin\theta }/2$ \Rightarrow tan $\theta /2$ = n

The position vectors $\vec{OA} = \sqrt{3}\hat{i}+\hat{j}$ and $\vec{OB} = \hat{i}+\sqrt{3}\hat{j}$ are symmetric about the line $x-y=0$, which functions as the bisector of the acute angle between them. Point $C$ follows the vector $\beta\hat{i}+(1-\beta)\hat{j}$, corresponding to the coordinates $(\beta, 1-\beta)$. The perpendicular distance from this point to the line $x-y=0$ equals $\frac{3}{\sqrt{2}}$. Using the distance formula $\frac{|ax_0+by_0+c|}{\sqrt{a^{2+}b^2}}$, the relationship between these values is established as:

$\frac{|\beta-(1-\beta)|}{\sqrt{1^{2+}(-1)^2}} = \frac{3}{\sqrt{2}}$

Simplifying the expression results in $\frac{|2\beta-1|}{\sqrt{2}} = \frac{3}{\sqrt{2}}$, which implies $|2\beta-1|=3$. Solving this absolute value equation yields two cases: $2\beta-1=3$, giving $\beta=2$, and $2\beta-1=-3$, giving $\beta=-1$. Adding these results produces a total sum of $1$.

Equality between vectors requires that each corresponding scalar component must be identical, while the condition of perpendicularity is satisfied when the dot product of two vectors is zero. Given $\vec{b} = 2\vec{a}$, equating the coefficients of the unit vectors shows that the $\hat{j}$ components must satisfy the relation $3 - \lambda_2 = 2\lambda_1$, which can be rearranged as $2\lambda_1 + \lambda_2 = 3$.

For the perpendicular condition $\vec{a} \cdot \vec{c} = 0$, we calculate the sum of the products of corresponding components:
$(2)(3) + (6)(\lambda_1) + (3)(\lambda_3 - 1) = 0$
Expanding this expression gives $6 + 6\lambda_1 + 3\lambda_3 - 3 = 0$, which simplifies to $3 + 6\lambda_1 + 3\lambda_3 = 0$. Dividing the entire equation by 3 yields $2\lambda_1 + \lambda_3 = -1$. Substituting $\lambda_1 = -\frac{1}{2}$ into these derived equations confirms that $\lambda_2 = 4$ and $\lambda_3 = 0$.

A unit vector is defined by its direction cosines, where the sum of the squares of the cosines of the angles it makes with the coordinate axes must equal 1. This is expressed mathematically as $\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$.

By substituting the angles $\frac{\pi}{3}$ and $\frac{\pi}{4}$ into this relationship, the equation becomes $\cos^2(\frac{\pi}{3}) + \cos^2(\frac{\pi}{4}) + \cos^2 \theta = 1$. Evaluating the squares of these trigonometric values gives $\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$, which simplifies to $\frac{3}{4} + \cos^2 \theta = 1$. Subtracting $\frac{3}{4}$ from both sides leaves $\cos^2 \theta = \frac{1}{4}$, which implies $\cos \theta = \pm\frac{1}{2}$. Within the interval $(0, \pi)$, the possible values for $\theta$ are $\frac{\pi}{3}$ and $\frac{2\pi}{3}$.