Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

Since the vector $\vec{a}$ lies in the plane of $\vec{b}$ and $\vec{c}$, it can be expressed as a linear combination $\vec{a} = \lambda\vec{b} + \mu\vec{c}$. Substituting the given vectors, we obtain $\vec{a} = \lambda(2\hat{i}+\hat{j}+\hat{k}) + \mu(\hat{i}-\hat{j}+\hat{k}) = (2\lambda+\mu)\hat{i} + (\lambda-\mu)\hat{j} + (\lambda+\mu)\hat{k}$. Because $\vec{a}$ is perpendicular to $\vec{d}=3\hat{i}+2\hat{j}+6\hat{k}$, their dot product must be zero, which leads to $3(2\lambda+\mu) + 2(\lambda-\mu) + 6(\lambda+\mu) = 14\lambda + 7\mu = 0$, or $\mu = -2\lambda$. Substituting $\mu = -2\lambda$ into our expression for $\vec{a}$ yields $\vec{a} = 0\hat{i} + 3\lambda\hat{j} - \lambda\hat{k}$.

Applying the magnitude condition $|\vec{a}| = \sqrt{10}$, we find $\sqrt{(3\lambda)^2 + (-\lambda)^2} = \sqrt{9\lambda^{2+}\lambda^2} = \sqrt{10\lambda^2} = \sqrt{10}|\lambda| = \sqrt{10}$, implying that $|\lambda|=1$. We are asked to evaluate $[\vec{a}\vec{b}\vec{c}]+[\vec{a}\vec{b}\vec{d}]+[\vec{a}\vec{c}\vec{d}]$. Note that the first term $[\vec{a}\vec{b}\vec{c}]$ is zero because $\vec{a}, \vec{b},$ and $\vec{c}$ are coplanar. Using the linearity property of the scalar triple product, the sum of the remaining terms simplifies to $[\vec{a} \ (\vec{b}+\vec{c}) \ \vec{d}]$.

Given $\vec{b}+\vec{c} = (2+1)\hat{i} + (1-1)\hat{j} + (1+1)\hat{k} = 3\hat{i} + 0\hat{j} + 2\hat{k}$, the required value is the determinant of the matrix formed by $\vec{a}, \vec{b}+\vec{c},$ and $\vec{d}$:
$\begin{vmatrix} 0 & 3\lambda & -\lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6 \end{vmatrix}$
Expanding along the first row, we get $-3\lambda(18-6) - \lambda(6-0) = -36\lambda - 6\lambda = -42\lambda$. Taking $\lambda=1$, we arrive at the value $-42$.

The projection of a vector $\vec{a}$ onto a vector $\vec{s}$ is defined as $\frac{\vec{a} \cdot \vec{s}}{|\vec{s}|}$. With $\vec{a} = \hat{i}+2\hat{j}+\hat{k}$ and the sum vector $\vec{s} = (2-\lambda)\hat{i}+6\hat{j}-2\hat{k}$, the condition requires that $\frac{\vec{a} \cdot \vec{s}}{|\vec{s}|} = 1$, which implies $\vec{a} \cdot \vec{s} = |\vec{s}|$.

The dot product equals $(1)(2-\lambda) + (2)(6) + (1)(-2)$, which simplifies to $12 - \lambda$. The magnitude of the sum vector $|\vec{s}|$ corresponds to $\sqrt{(2-\lambda)^2 + 6^2 + (-2)^2} = \sqrt{(2-\lambda)^2 + 40}$. Setting the dot product equal to the magnitude yields $12 - \lambda = \sqrt{(2-\lambda)^2 + 40}$, which becomes $(12-\lambda)^2 = (2-\lambda)^2 + 40$ upon squaring both sides. Expanding these terms results in $144 - 24\lambda + \lambda^2 = 4 - 4\lambda + \lambda^2 + 40$. Simplifying this equation leaves $100 = 20\lambda$, establishing that $\lambda = 5$.

From given figure, we can write all the forces in vector notation as {\text{OP}}=(20 \sin 30° {{i}}^{\^}+20 \cos 30° {{j}}^{\^}) {\text{N}} =(10 {{i}}^{\^}+17.32 {{j}}^{\^}) {\text{N}} {\text{OQ}}=(10 \cos 30° {{i}}^{\^}+10 \sin 30° {{j}}^{\^}) {\text{N}} =(8.66 {{i}}^{\^}+5 {{j}}^{\^}) {\text{N}} {\text{OR}}=(20 \cos 45° {{i}}^{\^}-20 \sin 45° {{j}}^{\^}) {\text{N}} =(14.14 {{i}}^{\^}-14.14 {{j}}^{\^}) {\text{N}} {\text{OS}}=(-15 \cos 45° {{i}}^{\^}-15 \sin 45° {{j}}^{\^}) {\text{N}} =(-10.60 {{i}}^{\^}-10.60 {{j}}^{\^}) {\text{N}} {\text{OT}}=(-15 \sin 60°\hat{i}+15 \cos 60°{{j}}^{\^}) {\text{N}} =(-12.99 {{i}}^{\^}+7.5 {{j}}^{\^}) {\text{N}} The resultant of these forces is given as F = OP + OQ + OR + OS + OT {\text{F}}=[(10 {{i}}^{\^}+17.32 {{j}}^{\^})+(8.66 {{i}}^{\^}+5 {{j}}^{\^}) +(14.14 {{i}}^{\^}-14.14 {{i}}^{\^})+(-10.60 {{i}}^{\^}-10.60 {{j}}^{\^}) +(-12.99 {{i}}^{\^}+7.5 {{j}}^{\^})] {\text{F}}=9.21 {{i}}^{\^}+5.08 {{j}}^{\^} {\text{N}} {\text{F}} \approx 9.25 {{i}}^{\^}+5 {{j}}^{\^} {\text{N}} Thus, the resultant force is {\text{F}}\approx 9.25 {{i}}^{\^}+5 {{j}}^{\^} {\text{N}}

Given, $P \times Q=Q \times P$ $\Rightarrow \;\; P \times Q=-P \times Q \Rightarrow 2(P \times Q)=0$ or $P \times Q=0$ It is possible only, if $P=0$ or $Q=0$, angle between them is $180^{\circ}$. $\therefore \;\; \theta =180^{\circ}$

When a vector $\overset{\rightarrow}{c}$ is perpendicular to two given vectors $\overset{\rightarrow}{a}$ and $\overset{\rightarrow}{b}$, it must be parallel to their cross product, expressible as $\overset{\rightarrow}{c} = \lambda(\overset{\rightarrow}{a} \times \overset{\rightarrow}{b})$. The cross product of $\overset{\rightarrow}{a} = \hat{i}+\hat{j}-\hat{k}$ and $\overset{\rightarrow}{b} = \hat{i}+2\hat{j}+\hat{k}$ is found by computing the determinant:
$\overset{\rightarrow}{a} \times \overset{\rightarrow}{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -1 \\ 1 & 2 & 1 \end{vmatrix} = 3\hat{i}-2\hat{j}+\hat{k}$
Using the condition $\overset{\rightarrow}{c} \cdot (\hat{i}+\hat{j}+3\hat{k}) = 8$, we substitute the expression for $\overset{\rightarrow}{c}$ to find $\lambda(3\hat{i}-2\hat{j}+\hat{k}) \cdot (\hat{i}+\hat{j}+3\hat{k}) = 8$, which simplifies to $4\lambda = 8$, yielding $\lambda = 2$. Finally, we calculate the required value as $\overset{\rightarrow}{c} \cdot (\overset{\rightarrow}{a} \times \overset{\rightarrow}{b}) = 2(3\hat{i}-2\hat{j}+\hat{k}) \cdot (3\hat{i}-2\hat{j}+\hat{k}) = 2(9+4+1)$, which equals 28.

To determine the unknown variables $\alpha$ and $\beta$, we start by evaluating the provided dot products. For the vectors $a = \langle \alpha, \beta, 3 \rangle$ and $b = \langle -\beta, -\alpha, -1 \rangle$, the condition $a \cdot b = 1$ expands to $-2\alpha\beta - 3 = 1$, which simplifies to the product $\alpha\beta = -2$. Utilizing the vector $c = \langle 1, -2, -1 \rangle$, the requirement $b \cdot c = -3$ yields the linear equation $-\beta + 2\alpha + 1 = -3$, which rearranges to $\beta - 2\alpha = 4$.

Substituting $\beta = 2\alpha + 4$ into the product equation gives $\alpha(2\alpha + 4) = -2$, resulting in the quadratic expression $\alpha^2 + 2\alpha + 1 = 0$. This factors as $(\alpha + 1)^2 = 0$, confirming that $\alpha = -1$, which leads to $\beta = 2$. Consequently, the vectors are defined as $a = \langle -1, 2, 3 \rangle$, $b = \langle -2, 1, -1 \rangle$, and $c = \langle 1, -2, -1 \rangle$.

The expression $\frac{1}{3}[(a \times b) \cdot c]$ utilizes the scalar triple product, which corresponds to the determinant of the matrix formed by the components of these vectors:
$\begin{vmatrix} -1 & 2 & 3 \\ -2 & 1 & -1 \\ 1 & -2 & -1 \end{vmatrix}$
Expanding this determinant yields $-1(-1 - 2) - 2(2 + 1) + 3(4 - 1)$, which simplifies to $3 - 6 + 9 = 6$. Dividing this result by 3 gives the final value of 2.

Rotation about the origin preserves the magnitude of a vector, so equating the squares of the lengths of $\vec{v}_{1}=\sqrt{3}p\hat{i}+\hat{j}$ and $\vec{v}_{2}=2\hat{i}+(p+1)\hat{j}$ gives
$3p^{2+}1 = 4+(p+1)^2$
Simplifying this expression leads to the quadratic $2p^{2-}2p-4=0$, which factors as $2(p-2)(p+1)=0$. Given the condition $p>0$, we must have $p=2$, which yields the vectors $\vec{v}_{1}=2\sqrt{3}\hat{i}+\hat{j}$ and $\vec{v}_{2}=2\hat{i}+3\hat{j}$, both having a magnitude of $\sqrt{13}$.

Using the definition of the dot product $\vec{v}_{1}\cdot\vec{v}_{2}=|\vec{v}_{1}||\vec{v}_{2}|\cos\theta$, we find
$\cos\theta = \frac{(2\sqrt{3})(2)+(1)(3)}{\sqrt{13}\sqrt{13}} = \frac{4\sqrt{3}+3}{13}$
From this, we determine the sine of the rotation angle using $\sin\theta = \sqrt{1-\cos^2\theta} = \frac{\sqrt{112-24\sqrt{3}}}{13}$. Noting that the square of $6\sqrt{3}-2$ is exactly $112-24\sqrt{3}$, we express the tangent as $\tan\theta = \frac{6\sqrt{3}-2}{4\sqrt{3}+3}$. Comparing this result to the provided form $\frac{\alpha\sqrt{3}-2}{4\sqrt{3}+3}$ confirms that $\alpha=6$.

The cross product condition $r \times a = r \times b$ implies $r \times (a - b) = 0$, meaning the vector $r$ is parallel to the difference vector $(a - b)$. Performing the subtraction between $a = 2\hat{i} - 3\hat{j} + 4\hat{k}$ and $b = 7\hat{i} + \hat{j} - 6\hat{k}$ yields $-5\hat{i} - 4\hat{j} + 10\hat{k}$, so we represent $r$ as the scalar multiple $\lambda(-5\hat{i} - 4\hat{j} + 10\hat{k})$.

Solving the condition $r \cdot (\hat{i} + 2\hat{j} + \hat{k}) = -3$ for the scalar $\lambda$ yields:
$\lambda(-5\hat{i} - 4\hat{j} + 10\hat{k}) \cdot (\hat{i} + 2\hat{j} + \hat{k}) = \lambda(-5 - 8 + 10) = -3\lambda = -3$
This confirms $\lambda = 1$, giving $r = -5\hat{i} - 4\hat{j} + 10\hat{k}$. The final required dot product is calculated as:
$r \cdot (2\hat{i} - 3\hat{j} + \hat{k}) = (-5)(2) + (-4)(-3) + (10)(1) = 12$

The scalar triple product represents the volume of a parallelepiped defined by three vectors and is computed as the determinant of a matrix formed by their components. To find this value, we must first determine the unknown integers $\alpha$ and $\beta$ by applying the provided dot product constraints.

For the dot product $\vec{a} \cdot \vec{b} = -1$, we multiply corresponding components to get $(1)(3) + (-\alpha)(\beta) + (\beta)(-\alpha) = -1$, which simplifies to $3 - 2\alpha\beta = -1$, or $\alpha\beta = 2$. Similarly, using the condition $\vec{b} \cdot \vec{c} = 10$, we calculate $(3)(-\alpha) + (\beta)(-2) + (-\alpha)(1) = 10$, leading to $-4\alpha - 2\beta = 10$, which reduces to $2\alpha + \beta = -5$. Since $\alpha$ and $\beta$ are integers, we can express $\beta$ as $-5 - 2\alpha$ and substitute this into $\alpha\beta = 2$ to form the quadratic equation $2\alpha^2 + 5\alpha + 2 = 0$. Solving this yields $\alpha = -2$, and substituting this back gives $\beta = -1$.

With the integer values established, the vectors are $\vec{a} = (1, 2, -1)$, $\vec{b} = (3, -1, 2)$, and $\vec{c} = (2, -2, 1)$. We evaluate the scalar triple product $(\vec{a} \times \vec{b}) \cdot \vec{c}$ by calculating the determinant:

Expanding this determinant yields $1((-1)(1) - (2)(-2)) - 2((3)(1) - (2)(2)) - 1((3)(-2) - (-1)(2))$, which simplifies to $1(3) - 2(-1) - 1(-4)$. Summing these results gives $3 + 2 + 4 = 9$.

(A \cos \theta ) {\hat{B}}=A(\frac{\vec{A} \.\vec{B}}{A B}) \hat{B}=\frac{{A} ^{\to }\.{B} ^{\to }}{B} \hat{B} $=\frac{2}{\\sqrt{2}}(\frac{\hat{i}+\hat{j}}{\\sqrt{2}})=\hat{i}+\hat{j}$

The scalar projection of $\overrightarrow{AB}$ onto $\overrightarrow{AC}$ is equivalent to the length of the segment formed by dropping a perpendicular from $B$ to the line containing $AC$, which is calculated as $|\overrightarrow{AB}| \cos \theta$, where $\theta$ is the angle at vertex $A$. Applying the Law of Cosines to triangle $ABC$ with side lengths $|\overrightarrow{AB}|=10$, $|\overrightarrow{AC}|=7$, and $|\overrightarrow{BC}|=8$, we obtain the relationship $8^2 = 10^2 + 7^2 - 2(10)(7) \cos \theta$. Simplifying the squares and products results \in the equation $64 = 149 - 140 \cos \theta$, which simplifies further to $140 \cos \theta = 85$. Solving for $\cos \theta$ gives $\frac{85}{140}$, and multiplying this value by the magnitude $|\overrightarrow{AB}|=10$ yields the final projection of $10 \times \frac{85}{140} = \frac{85}{14}$.

When the cross product of a vector $r$ with one vector is equal to the cross product of a second vector with $r$, the expression $r \times a = b \times r$ can be rearranged to $r \times a + r \times b = 0$, which simplifies to $r \times (a + b) = 0$. This mathematical condition indicates that $r$ must be collinear with the sum of the two vectors. Adding the given vectors $a = \hat{i} + 2\hat{j} - 3\hat{k}$ and $b = 2\hat{i} - 3\hat{j} + 5\hat{k}$ results in $a + b = 3\hat{i} - \hat{j} + 2\hat{k}$. Therefore, $r$ can be defined as $r = \lambda(3\hat{i} - \hat{j} + 2\hat{k})$ for some scalar $\lambda$.

Applying this relationship to the first dot product $r \cdot (\alpha \hat{i} + 2 \hat{j} + \hat{k}) = 3$ gives the equation $\lambda(3\alpha - 2 + 2) = 3$, which simplifies to $3\alpha\lambda = 3$, or $\alpha\lambda = 1$. Utilizing the second dot product $r \cdot (2 \hat{i} + 5 \hat{j} - \alpha \hat{k}) = -1$ leads to $\lambda(6 - 5 - 2\alpha) = -1$. Expanding this expression provides $\lambda(1 - 2\alpha) = -1$, which is equivalent to $\lambda - 2\alpha\lambda = -1$. By substituting the previously derived $\alpha\lambda = 1$ into this equation, we find $\lambda - 2(1) = -1$, confirming that $\lambda = 1$. Given that $\alpha\lambda = 1$ and $\lambda = 1$, it follows that $\alpha = 1$.

With $\lambda$ determined, the vector becomes $r = 3\hat{i} - \hat{j} + 2\hat{k}$. The squared magnitude is calculated as $|r|^2 = 3^2 + (-1)^2 + 2^2 = 9 + 1 + 4 = 14$. Combining these findings, the value of $\alpha + |r|^2$ is $1 + 14 = 15$.

The rotation of a coordinate system about the origin acts as a rigid transformation, meaning the magnitude of any vector remains constant regardless of its representation in the new basis. By equating the squared norm of the original vector components, $(3p)^2 + 1^2$, with the squared norm of the transformed components, $(p+1)^2 + (\sqrt{10})^2$, we can determine the valid value for $p$.

Expanding these squares leads to the following relationship:
$9p^2 + 1 = p^2 + 2p + 1 + 10$
This simplifies to $8p^2 - 2p - 10 = 0$, which can be reduced further to $4p^2 - p - 5 = 0$. Factoring this expression yields $(4p - 5)(p + 1) = 0$, providing two potential values for $p$, which are $\frac{5}{4}$ and $-1$. Among the available options, only $-1$ matches these results.

The geometric definitions $\vec{a} \times \vec{b} = \vec{c}$ and $\vec{b} \times \vec{c} = \vec{a}$ imply that the vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are mutually orthogonal. By calculating magnitudes, the relation $|\vec{a} \times \vec{b}| = |\vec{c}|$ leads to $|\vec{a}| |\vec{b}| = |\vec{c}|$, and $|\vec{b} \times \vec{c}| = |\vec{a}|$ leads to $|\vec{b}| |\vec{c}| = |\vec{a}|$. Given $|\vec{a}| = 2$, the first equation becomes $2 |\vec{b}| = |\vec{c}|$. Substituting this into the second equation gives $|\vec{b}| (2 |\vec{b}|) = 2$, which simplifies to $2 |\vec{b}|^2 = 2$, meaning $|\vec{b}| = 1$ and $|\vec{c}| = 2$.

The expression $\vec{a} \times ((\vec{b} + \vec{c}) \times (\vec{b} - \vec{c}))$ expands to $\vec{a} \times (\vec{b} \times \vec{b} - \vec{b} \times \vec{c} + \vec{c} \times \vec{b} - \vec{c} \times \vec{c})$. Since the cross product of any vector with itself is the zero vector, this becomes $\vec{a} \times (-\vec{b} \times \vec{c} - \vec{b} \times \vec{c}) = \vec{a} \times (-2 \vec{a})$, which is indeed $\vec{0}$. The projection of $\vec{a}$ onto $\vec{b} \times \vec{c}$ is calculated as $\frac{\vec{a} \cdot (\vec{b} \times \vec{c})}{|\vec{b} \times \vec{c}|}$. Substituting $\vec{b} \times \vec{c} = \vec{a}$, the expression becomes $\frac{\vec{a} \cdot \vec{a}}{|\vec{a}|} = \frac{|\vec{a}|^2}{|\vec{a}|} = |\vec{a}| = 2$.

Regarding the scalar triple products,

is equal to $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{a} \cdot \vec{a} = 4$. Similarly,

is $\vec{c} \cdot (\vec{a} \times \vec{b}) = \vec{c} \cdot \vec{c} = 4$. Their sum is $4 + 4 = 8$. Finally, the squared magnitude $|3 \vec{a} + \vec{b} - 2 \vec{c}|^2$ expands to $9|\vec{a}|^2 + |\vec{b}|^2 + 4|\vec{c}|^2$ because the dot products of distinct orthogonal vectors are zero. Substituting the known magnitudes results in $9(4) + 1 + 4(4) = 36 + 1 + 16 = 53$, which means the value of $51$ is incorrect.

The relationship between the diagonal vectors and the height of a space is governed by the scalar product formula, which connects the angle between these vectors to the geometric dimensions of the hall. By setting the corner $A$ at the origin $(0,0,0)$ of a 3D coordinate system, we define the vertices $B=(10,0,0)$, $G=(10,10,h)$, and $H=(0,10,h)$, where $h$ is the unknown height. The vectors representing the diagonals are then $\vec{AG} = 10\hat{i} + 10\hat{j} + h\hat{k}$ and $\vec{BH} = -10\hat{i} + 10\hat{j} + h\hat{k}$.

Applying the definition $\cos \theta = \frac{\vec{AG} \cdot \vec{BH}}{|\vec{AG}| |\vec{BH}|}$ with the given value $\cos \theta = \frac{1}{5}$, we compute the dot product $\vec{AG} \cdot \vec{BH} = -100 + 100 + h^2 = h^2$ and the magnitude of each vector $|\vec{AG}| = |\vec{BH}| = \sqrt{10^2 + 10^2 + h^2} = \sqrt{200 + h^2}$. This establishes the following equation:
$\frac{1}{5} = \frac{h^2}{200 + h^2}$
Rearranging the terms yields $200 + h^2 = 5h^2$, which simplifies to $4h^2 = 200$. Solving for $h$ gives $h^2 = 50$, resulting in a height of $5\sqrt{2}\,\mathrm{m}$.

Given ${X}={Y}$ $\\sqrt{X^{2}+Y^{2}-2 \\times Y \cos \theta }$ $=n \\sqrt{X^{2}+Y^{2}+2 \\times Y \cos \theta }$ Square both sides $2 X^{2}(1-\cos \theta )=n^{2} \.2 X^{2}(1+\cos \theta )$ $1-\cos \theta =n^{2}+n^{2} \cos \theta$ $\cos \theta =\frac{1-{n}^{2}}{1+{n}^{2}}$ $\theta =\cos ^{-1}[\frac{n^{2}-1}{-n^{2}-1}]$

Collinear vectors share proportional components, so given $a_1=x\hat{i}-\hat{j}+\hat{k}$ and $a_2=\hat{i}+y\hat{j}+z\hat{k}$, we identify the ratios $\frac{x}{1}=\frac{-1}{y}=\frac{1}{z}=\lambda$, which leads to $x=\lambda$ and $z=\frac{1}{\lambda}$. For the vector $x\hat{i}+\hat{j}+z\hat{k}$, we substitute these values to get $\lambda\hat{i}+\hat{j}+\frac{1}{\lambda}\hat{k}$, and by assigning $\lambda=-1$ to match the $\hat{j}$ coefficient, we obtain $-\hat{i}+\hat{j}-\hat{k}$. The magnitude of this vector is $\sqrt{(-1)^{2+}1^{2+}(-1)^2}=\sqrt{3}$, so the normalized unit vector is $\frac{1}{\sqrt{3}}(-\hat{i}+\hat{j}-\hat{k})$. Since a vector parallel to this may have the opposite direction, the form $\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}+\hat{k})$ satisfies the condition.

Since vector $c$ lies in the plane spanned by $a$ and $b$ and is also perpendicular to $b$, we can express it using the vector triple product formula as $c = \lambda [b \times (a \times b)]$, which simplifies to $c = \lambda [(b \cdot b) a - (b \cdot a) b]$. Given the vectors $a = -\hat{i} + \hat{j} + \hat{k}$ and $b = 2\hat{i} + \hat{k}$, we first compute the necessary dot products: $|b|^2 = 2^2 + 0^2 + 1^2 = 5$ and $a \cdot b = (-1)(2) + (1)(0) + (1)(1) = -1$. Substituting these values into our expression for $c$, we obtain $c = \lambda [5a - (-1)b] = \lambda (5a + b)$.