Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

For two perpendicular vectors$\;(a {i}^{\wedge }+b {j}^{\wedge }+{k}^{\wedge }) \cdot (2 {i}^{\wedge }-3 {j}^{\wedge }+4 {k}^{\wedge })=0$ $\;2 a-3 b+4=0$On solving, $2 a-3 b=-4$Also given$3 a+2 b=7$We get ${\text{a}}=1, {\text{b}}=2$ $\;\;{ {\text{a}}}/{ {\text{b}}}=\;{ {\text{x}}}/{2} \Rightarrow {\text{x}}=\;{2 {\text{a}}}/{ {\text{b}}}=\;\frac{2 \times 1}{2}$ $\Rightarrow {\text{x}}=1$

$\; {\text{A}}_{ {\text{y}}}= {\text{A}} \cos 30^{\circ}=2 {\sqrt{3}}$ $\; {\text{A}}=4$ $\; {\text{A}}_{ {\text{x}}}= {\text{A}} sin \;30^{\circ}=2$

When four points lie on the same plane, the scalar triple product of the vectors formed by connecting them from a single common point must be zero. By designating the points as $A(1, -2, 3)$, $B(2, -3, 4)$, $C(\alpha+1, 0, 2)$, and $D(9, \alpha-8, 6)$, we can define three vectors originating from point $A$: $\vec{AB} = \hat{i} - \hat{j} + \hat{k}$, $\vec{AC} = \alpha\hat{i} + 2\hat{j} - \hat{k}$, and $\vec{AD} = 8\hat{i} + (\alpha-6)\hat{j} + 3\hat{k}$. Setting the determinant of these components to zero establishes the condition for coplanarity:

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Expanding this determinant along the first row leads to the expression $1(6 + \alpha - 6) + 1(3\alpha + 8) + 1(\alpha^2 - 6\alpha - 16) = 0$, which simplifies to the quadratic equation $\alpha^2 - 2\alpha - 8 = 0$. Solving this equation gives the roots $\alpha = 4$ and $\alpha = -2$. The sum of all possible values for $\alpha$ is therefore $4 + (-2) = 2$.

The vector triple product identity provides the most direct path to connecting these vector quantities. First, recognize that the dot product $\vec{a} \cdot \vec{b} = (1)(1) + (2)(1) + (3)(-1) = 0$, which implies that the vectors $\vec{a}$ and $\vec{b}$ are orthogonal. Given the condition $\vec{b} \cdot \vec{c} = -\sqrt{3}|\vec{b}|$, and calculating $|\vec{b}| = \sqrt{1^{2+}1^{2+}(-1)^2} = \sqrt{3}$, we determine that $\vec{b} \cdot \vec{c} = -3$. Applying the identity $\vec{b} \times (\vec{a} \times \vec{c}) = (\vec{b} \cdot \vec{c})\vec{a} - (\vec{b} \cdot \vec{a})\vec{c}$, the orthogonality reduces this expression to $\vec{b} \times (\vec{a} \times \vec{c}) = -3\vec{a}$. Squaring the magnitude of both sides results in $|\vec{b} \times (\vec{a} \times \vec{c})|^2 = 9|\vec{a}|^2 = 9(1^{2+}2^{2+}3^2) = 126$. Utilizing the identity $|\vec{b} \times \vec{v}|^2 = |\vec{b}|^2|\vec{v}|^2 - (\vec{b} \cdot \vec{v})^2$ where $\vec{v} = \vec{a} \times \vec{c}$, we substitute $126 = 3|\vec{a} \times \vec{c}|^2 - (27)^2$. This leads to $3|\vec{a} \times \vec{c}|^2 = 126 + 729 = 855$, which simplifies to $|\vec{a} \times \vec{c}|^2 = 285$.

The cross product $\vec{b} \times \vec{c}$ results in a vector that must be perpendicular to both $\vec{b}$ and $\vec{c}$, meaning the dot product $\vec{a} \cdot \vec{c}$ must equal zero if $\vec{b} \times \vec{c} = \vec{a}$. We evaluate this condition by computing the dot product of $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{c} = 2\hat{i} - 3\hat{j} + 2\hat{k}$. Calculating the sum of the products of corresponding components gives $(1)(2) + (1)(-3) + (-1)(2) = 2 - 3 - 2 = -3$. Since this result is not equal to zero, no vector $\vec{b}$ can satisfy the given equation, resulting in zero possible vectors.

The condition for a vector $\overset{\rightarrow}{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$ to be perpendicular to $3 \hat{i} + \frac{1}{2} \hat{j} + 2 \hat{k}$ is that their dot product equals zero, yielding $3 a_1 + \frac{1}{2} a_2 + 2 a_3 = 0$. Simultaneously, the cross product $\overset{\rightarrow}{a} \times (2 \hat{i} + \hat{k})$ is determined by evaluating the determinant of the matrix with unit vectors in the first row, $(a_1, a_2, a_3)$ in the second, and $(2, 0, 1)$ in the third, which results in $a_2 \hat{i} + (2 a_3 - a_1) \hat{j} - 2 a_2 \hat{k}$.

Equating this expression to $2 \hat{i} - 13 \hat{j} - 4 \hat{k}$ allows us to identify $a_2 = 2$ directly from the $\hat{i}$ components, a value that is consistent with the $\hat{k}$ component $-2 a_2 = -4$. Substituting $a_2 = 2$ into the dot product equation gives $3 a_1 + 1 + 2 a_3 = 0$, or $3 a_1 + 2 a_3 = -1$. From the $\hat{j}$ component of the cross product, we have $-(a_1 - 2 a_3) = -13$, which simplifies to $a_1 - 2 a_3 = 13$. Adding these two linear equations, $(3 a_1 + 2 a_3) + (a_1 - 2 a_3) = -1 + 13$, results in $4 a_1 = 12$, identifying $a_1 = 3$, which then determines $a_3 = -5$.

With the vector established as $\overset{\rightarrow}{a} = 3 \hat{i} + 2 \hat{j} - 5 \hat{k}$, the projection of this vector onto $2 \hat{i} + 2 \hat{j} + \hat{k}$ is calculated by the dot product of the two vectors divided by the magnitude of the target vector. Computing the dot product gives $(3)(2) + (2)(2) + (-5)(1) = 6 + 4 - 5 = 5$, while the magnitude of the target vector is $\sqrt{2^2 + 2^2 + 1^2} = 3$, leading to a final projection value of $\frac{5}{3}$.

{ \vec{\text{a}} \cdot { \vec{\text{b}}=0 \therefore { \vec{\text{a}} \cdot { \vec{\text{b}}=0 $\therefore 2 \times 1+4 \times 2-2 \times \alpha =0$ $\therefore \alpha =5$

The vector triple product identity $(\vec{A} \times \vec{B}) \times \vec{C} = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{B} \cdot \vec{C})\vec{A}$ simplifies the given expression $((\vec{a} \times \vec{b}) \times \hat{i}) \cdot \hat{k} = \frac{23}{2}$ by expanding it into $((\vec{a} \cdot \hat{i})\vec{b} - (\vec{b} \cdot \hat{i})\vec{a}) \cdot \hat{k} = \frac{23}{2}$. Given $\vec{a} \cdot \hat{i} = 2$ and $\vec{b} \cdot \hat{i} = \alpha$, this expression becomes $(2\vec{b} - \alpha\vec{a}) \cdot \hat{k} = \frac{23}{2}$. Since $\vec{b} \cdot \hat{k} = 2$ and $\vec{a} \cdot \hat{k} = 5$, substituting these values yields $2(2) - \alpha(5) = \frac{23}{2}$, which simplifies to $4 - 5\alpha = \frac{23}{2}$, resulting in $\alpha = -\frac{3}{2}$.

To determine the magnitude $|\vec{b} \times 2\hat{j}|$, note that it equals $2|\vec{b} \times \hat{j}|$. Calculating the cross product $\vec{b} \times \hat{j}$ involves $(\alpha \hat{i} + \beta \hat{j} + 2\hat{k}) \times \hat{j} = \alpha \hat{k} - 2\hat{i}$, so the expression becomes $|2(\alpha \hat{k} - 2\hat{i})| = |2\alpha \hat{k} - 4\hat{i}|$. The magnitude is $\sqrt{(2\alpha)^2 + (-4)^2} = \sqrt{4\alpha^2 + 16}$. Substituting $\alpha = -\frac{3}{2}$ provides $\sqrt{4(-\frac{3}{2})^2 + 16} = \sqrt{4(\frac{9}{4}) + 16} = \sqrt{9 + 16} = 5$.

The scalar triple product of three vectors is zero whenever those vectors are coplanar, as this implies that one vector can be expressed as a linear combination of the other two. Expanding each term using the vector triple product identity $\vec{a}\times (\vec{b}\times \vec{c}) = (\vec{a}\cdot\vec{c})\vec{b} - (\vec{a}\cdot\vec{b})\vec{c}$ results in the expressions $3\vec{b} - \vec{c}$, $\vec{c} - 2\vec{a}$, and $3\vec{b} - 2\vec{a}$. Adding the first two expressions yields $3\vec{b} - 2\vec{a}$, confirming that the third vector is a linear combination of the first two. This linear dependence indicates that the vectors lie in the same plane, which makes their scalar triple product zero.

{ \vec{\text{A}}=(2 { {\text{i}}^{\^}+3 { {\text{j}}^{\^}-{ {\text{k}}^{\^}) {\text{m}} and { \vec{\text{B}}=({ {\text{i}}^{\^}+2 { {\text{j}}^{\^}+2 { {\text{k}}^{\^}) {\text{m}}Component of $\vec{A}$ along $\vec{B}=\vec{A} \cdot \hat{B}$ =\;{{ \vec{\text{A}} \cdot { \vec{\text{B}}}/{|{ \vec{\text{B}}|}=\;\frac{2+6-2}{\sqrt{1^{2+}2^{2+}2^2}} $=\;\frac{6}{3}=2$

The given condition $\vec{a}+(\vec{b}\times \vec{c})=\vec{0}$ implies that $\vec{a}=-(\vec{b}\times \vec{c})$, which means that the vector $\vec{a}$ is expressed as the negative cross product of $\vec{b}$ and $\vec{c}$. When we evaluate the expression $3(\vec{c}\cdot \vec{a})$ by substituting $\vec{a}$ with its equivalent, we obtain $3(\vec{c}\cdot (-(\vec{b}\times \vec{c})))$, or equivalently $-3(\vec{c}\cdot (\vec{b}\times \vec{c}))$. In this scalar triple product, because the vector $\vec{c}$ appears twice, the volume of the parallelepiped formed by these vectors is zero, forcing the entire expression to equal zero regardless of the specific components of the vectors involved. While this algebraic result is consistent with the vector properties, the specific vectors provided, $\vec{a}=\hat{i}-2\hat{j}+3\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}+\hat{k}$, yield a dot product of $\vec{a}\cdot \vec{b} = 1-2+3=2$, which contradicts the condition $\vec{a}\cdot \vec{b}=0$ required by the expression $\vec{a}=-(\vec{b}\times \vec{c})$, indicating that the problem statement contains an inherent numerical inconsistency.

The coplanarity of three vectors implies that their scalar triple product is zero, meaning the determinant of the matrix formed by their components must vanish. In addition, the perpendicularity condition $\vec{b} \cdot \vec{c} = 0$ and the known dot product $\vec{a} \cdot \vec{c} = 5$ provide specific linear constraints on the components of $\vec{c}$. From these dot product conditions, we derive the equations $2c_1 + c_2 + 3c_3 = 5$ and $3c_1 + 3c_2 + c_3 = 0$. The requirement for coplanarity involves evaluating the determinant formed by the components of $\vec{a}$, $\vec{b}$, and $\vec{c}$, which leads to the additional linear constraint $8c_1 - 7c_2 - 3c_3 = 0$.

By solving this system of three simultaneous linear equations, we find the individual values of the components to be $c_1 = \frac{10}{122}$, $c_2 = -\frac{85}{122}$, and $c_3 = \frac{225}{122}$. Summing these three components yields a value of $\frac{150}{122}$. Consequently, multiplying the sum by 122 results in $122(c_1 + c_2 + c_3) = 150$.

Since the vector cross product produces a result orthogonal to its factors, the expression $\hat{b} = \vec{c} + 2(\vec{c} \times \hat{a})$ simplifies upon calculating the dot product with $\vec{c}$, as the cross product term becomes zero, leaving $\hat{b} \cdot \vec{c} = |\vec{c}|^2$. Rearranging the original equation to $\hat{b} - \vec{c} = 2(\vec{c} \times \hat{a})$ and squaring both sides allows us to use the property $|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta$. The expansion $|\hat{b}|^2 + |\vec{c}|^2 - 2\hat{b} \cdot \vec{c} = 4|\vec{c}|^2|\hat{a}|^2 \sin^2(\frac{\pi}{12})$ then incorporates the unit vector magnitudes and the result $\hat{b} \cdot \vec{c} = |\vec{c}|^2$, leading to $1 + |\vec{c}|^2 - 2|\vec{c}|^2 = 4|\vec{c}|^2 \sin^2(\frac{\pi}{12})$. Using the identity $\sin^2(\frac{\pi}{12}) = \frac{2-\sqrt{3}}{4}$, the equation resolves to $1 - |\vec{c}|^2 = |\vec{c}|^2(2 - \sqrt{3})$, which further simplifies to $1 = |\vec{c}|^2(3 - \sqrt{3})$. Consequently, the squared magnitude $|6\vec{c}|^2 = 36|\vec{c}|^2$ is equivalent to $\frac{36}{3 - \sqrt{3}}$, which rationalizes to $6(3 + \sqrt{3})$.

$\vec{A} \times \vec{A}=A \times A \times \sin 0^{\circ}$ $=0$

$\because |\hat{A}-\hat{B}|=2 \sin (\;\frac{\theta }{2})$and, $|\hat{A}+\hat{B}|=2 \cos (\;\frac{\theta }{2})$ $\Rightarrow \;\frac{|\hat{A}-\hat{B}|}{|\hat{A}+\hat{B}|}=\tan (\;\frac{\theta }{2})$

${\sqrt{A^{2+}A^{2+}2 A^2 \cos \theta }}=2 {\sqrt{A^{2+}A^{2+}2 A^2(-\cos \theta )}}$ $\Rightarrow 2 A^{2+}2 A^2 \cos \theta =8 A^{2+}8 A^{2-}(-\cos \theta )$ $\Rightarrow 5 \cos \theta =3$ $\Rightarrow \theta =\cos ^{-1} (\;\frac{3}{5})$

The cosine of the angle $\theta$ between two vectors $\mathbf{u}$ and $\mathbf{v}$ is defined by the ratio $\cos \theta = \frac{\mathbf{u} \cdot \mathbf{v}}{|\mathbf{u}| |\mathbf{v}|}$. For unit vectors $\hat{a}$ and $\hat{b}$ with an angle of $\frac{\pi}{4}$ between them, their dot product is $\hat{a} \cdot \hat{b} = \cos \frac{\pi}{4} = \frac{1}{\sqrt{2}}$, and the magnitude of their cross product is $|\hat{a} \times \hat{b}| = \sin \frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Defining $\mathbf{u} = \hat{a} + \hat{b}$ and $\mathbf{v} = \hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})$, the dot product $\mathbf{u} \cdot \mathbf{v}$ expands to $(\hat{a} + \hat{b}) \cdot (\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b}))$. Since the cross product $\hat{a} \times \hat{b}$ is orthogonal to both $\hat{a}$ and $\hat{b}$, this expression simplifies to $|\hat{a}|^2 + 2\hat{b}\cdot\hat{b} + 2\hat{a}\cdot\hat{b} + \hat{b}\cdot\hat{a} = 1 + 2 + 3(\frac{1}{\sqrt{2}}) = 3 + \frac{3}{\sqrt{2}}$.

The magnitude squared of $\mathbf{u}$ is $|\hat{a}+\hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 + 2\hat{a}\cdot\hat{b} = 1 + 1 + 2(\frac{1}{\sqrt{2}}) = 2 + \sqrt{2}$. Similarly, the magnitude squared of $\mathbf{v}$ is $|\hat{a} + 2\hat{b} + 2(\hat{a} \times \hat{b})|^2 = |\hat{a}|^2 + |2\hat{b}|^2 + |2(\hat{a} \times \hat{b})|^2 + 2(\hat{a}\cdot 2\hat{b}) = 1 + 4 + 4(\frac{1}{2}) + 4(\frac{1}{\sqrt{2}}) = 7 + 2\sqrt{2}$, where cross-terms involving the cross product vanish due to orthogonality.

Substituting these values into $\cos^2 \theta = \frac{(\mathbf{u} \cdot \mathbf{v})^2}{|\mathbf{u}|^2 |\mathbf{v}|^2}$ results in $\cos^2 \theta = \frac{(3 + \frac{3}{\sqrt{2}})^2}{(2+\sqrt{2})(7+2\sqrt{2})}$. The numerator becomes $\frac{27+18\sqrt{2}}{2}$ and the denominator evaluates to $2(18+11\sqrt{2})$. Multiplying by $164$, the expression simplifies through rationalization to $164 \cos^2 \theta = 90 + 27\sqrt{2}$.

The projection of one vector onto another is defined by their dot product divided by the magnitude of the reference vector. To find this, first determine the cross product $\vec{a} \times \vec{b}$ using the determinant method:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 1 & -1 \\ 2 & 1 & -\alpha \end{vmatrix} = (1 - \alpha) \hat{i} + (\alpha^2 - 2) \hat{j} + (\alpha - 2) \hat{k}$
The magnitude of the target vector $-\hat{i} + 2\hat{j} - 2\hat{k}$ is $\sqrt{(-1)^2 + 2^2 + (-2)^2} = 3$. By applying the projection formula, set the ratio of the dot product to this magnitude equal to 30:
$\frac{|(1 - \alpha)(-1) + (\alpha^2 - 2)(2) + (\alpha - 2)(-2)|}{3} = 30$
Evaluating the numerator yields $|-1 + \alpha + 2\alpha^2 - 4 - 2\alpha + 4| = 90$, which simplifies to $|2\alpha^2 - \alpha - 1| = 90$. Solving the resulting quadratic equation $2\alpha^2 - \alpha - 91 = 0$ leads to the factors $(2\alpha + 13)(\alpha - 7) = 0$, giving potential roots of $\alpha = 7$ and $\alpha = -6.5$. Because the problem specifies $\alpha > 0$, the only valid result is $\alpha = 7$.

The relationship between the cross product and dot product of two vectors is governed by Lagrange's identity:
$|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$
For the given vector $\vec{a}=\hat{i}-\hat{j}+2\hat{k}$, the squared magnitude is $|\vec{a}|^2 = 1^{2+}(-1)^{2+}2^2 = 6$. Substituting the known values $|\vec{a} \times \vec{b}|^2 = 2^{2+}0^{2+}(-1)^2 = 5$ and $\vec{a} \cdot \vec{b} = 3$ into the identity gives $5 + 9 = 6|\vec{b}|^2$, which simplifies to $|\vec{b}|^2 = \frac{7}{3}$.

The projection of vector $\vec{b}$ onto $\vec{a}-\vec{b}$ is determined by the formula:
$\frac{\vec{b} \cdot (\vec{a}-\vec{b})}{|\vec{a}-\vec{b}|}$
We first calculate the denominator's magnitude using $|\vec{a}-\vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2(\vec{a} \cdot \vec{b}) = 6 + \frac{7}{3} - 6 = \frac{7}{3}$, so $|\vec{a}-\vec{b}| = \sqrt{\frac{7}{3}}$. Expanding the numerator gives $\vec{b} \cdot \vec{a} - |\vec{b}|^2$, which equals $3 - \frac{7}{3} = \frac{2}{3}$. Dividing the numerator by the denominator, we obtain $\frac{2/3}{\sqrt{7/3}} = \frac{2}{\sqrt{21}}$.

A vector making equal angles with the coordinate axes has equal direction cosines, defined by $\cos \alpha = \cos \beta = \cos \gamma = \frac{1}{\sqrt{3}}$. By representing the vector as $\vec{a} = k(\hat{i} + \hat{j} + \hat{k})$, we equate its projection on $3\hat{i} + 4\hat{j}$ to 7. This is determined via:
$\frac{k(3+4)}{\sqrt{3^{2+}4^2}} = \frac{7k}{5} = 7$
This calculation yields $k = 5$, establishing $\vec{a} = 5\hat{i} + 5\hat{j} + 5\hat{k}$.

Vector $\vec{b}$ is obtained by rotating $\vec{a}$ by $90^{\circ}$ within the plane containing $\vec{a}$ and the $x$-axis, ensuring $\vec{b}$ is orthogonal to $\vec{a}$ and coplanar with $\hat{i}$. Given these conditions, $\vec{b}$ takes the form $r(-2\hat{i} + \hat{j} + \hat{k})$. Since magnitude is preserved under rotation, we have $|\vec{b}| = |\vec{a}| = 5\sqrt{3}$, leading to the relation:
$|r|\sqrt{(-2)^2 + 1^2 + 1^2} = |r|\sqrt{6} = 5\sqrt{3}$
This results in $|r| = \frac{5}{\sqrt{2}}$.

The projection of $\vec{b}$ onto $3\hat{i} + 4\hat{j}$ is calculated as the absolute value of the dot product:
$\left| \frac{\vec{b} \cdot (3\hat{i} + 4\hat{j})}{\sqrt{3^{2+}4^2}} \right| = \left| \frac{r(-6 + 4)}{5} \right| = \frac{2|r|}{5}$
Substituting $|r| = \frac{5}{\sqrt{2}}$ into this expression provides a final projection length of $\sqrt{2}$.