Vector Algebra – JEE Main PhysicsPractice Questions & PYQs

When four points defined by the position vectors $\vec{a}, \vec{b}, \vec{c},$ and $\vec{d}$ are coplanar, the volume of the tetrahedron formed by these points is zero. This geometric condition implies that any three vectors constructed between these points, specifically $\vec{b}-\vec{a}$, $\vec{c}-\vec{b}$, and $\vec{d}-\vec{c}$, must lie in the same plane, causing their scalar triple product to vanish as $[\vec{b}-\vec{a}, \vec{c}-\vec{b}, \vec{d}-\vec{c}] = 0$. By expanding this scalar triple product according to the definition $(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{b}) \times (\vec{d}-\vec{c})) = 0$, we perform the vector algebra by distributing the dot and cross products. Many of the resulting scalar triple products contain repeated vectors, which simplifies those specific terms to zero. Through careful rearrangement of the remaining non-zero terms using the cyclic properties of scalar triple products, we successfully demonstrate that $[\vec{a}\vec{b}\vec{c}] = [\vec{d}\vec{c}\vec{a}] + [\vec{b}\vec{d}\vec{a}] + [\vec{c}\vec{d}\vec{b}]$.

$\;{| { {\text{F}}}^{\to }_1 |}= {\text{A}},{| { {\text{F}}}^{\to }_2 |}=\;{ {\text{A}}}/{2} \;\; \theta =\;\frac{\pi }{2}$ $\;{| { {\text{F}}}^{\to }_{\;\text{net }\;} |}=\sqrt{ {\text{F}}_1^{2+} {\text{F}}_2^2}$ $\;=\sqrt{ {\text{A}}^{2+} (\;{ {\text{A}}}/{2}) ^2}$ $\;{| { {\text{F}}}^{\to }_{\;\text{net }\;} |}=\;{{\sqrt{5}} {\text{A}}}/{2}$

When the cross product of a vector $\vec{r}$ with $\vec{b}$ is equal to the cross product of $\vec{c}$ with $\vec{b}$, it implies that the vector $\vec{r} - \vec{c}$ is parallel to $\vec{b}$. Rearranging $\vec{r} \times \vec{b} + \vec{b} \times \vec{c} = \vec{0}$ to $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ confirms that $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$, which leads to the linear relationship $\vec{r} = \vec{c} + \lambda \vec{b}$ for some scalar $\lambda$.

To determine $\lambda$, apply the orthogonality condition $\vec{r} \cdot \vec{a} = 0$. Substituting the expression for $\vec{r}$ into this equation yields $(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$, which expands to $\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$. Using the vectors $\vec{a} = \hat{i} + 2\hat{k}$, $\vec{b} = \hat{i} + \hat{j} + \hat{k}$, and $\vec{c} = 7\hat{i} - 3\hat{j} + 4\hat{k}$, the respective dot products are $\vec{c} \cdot \vec{a} = 15$ and $\vec{b} \cdot \vec{a} = 3$. Solving for $\lambda$ gives $\lambda = -15/3 = -5$.

The required value $\vec{r} \cdot \vec{c}$ is found by taking the dot product of $\vec{c}$ with the expression $\vec{c} + \lambda \vec{b}$, resulting in $|\vec{c}|^2 + \lambda (\vec{b} \cdot \vec{c})$. Substituting the values $|\vec{c}|^2 = 74$ and $\vec{b} \cdot \vec{c} = 8$ along with $\lambda = -5$ produces the final result of $74 - 5(8)$, which equals 34.

The magnitude of the cross product of two vectors is governed by the relationship $|\vec{a}| = |\vec{b}||\vec{c}|\sin\theta$, where $\theta$ is the angle between $\vec{b}$ and $\vec{c}$. Calculating the magnitudes of the provided vectors gives $|\vec{a}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$ and $|\vec{c}| = \sqrt{2^2 + (-3)^2 + 3^2} = \sqrt{22}$. Given $|\vec{b}|^2 = 50$, we have $|\vec{b}| = \sqrt{50}$, and substituting these into the cross product formula results in $\sqrt{11} = \sqrt{50} \cdot \sqrt{22} \cdot \sin\theta$. Simplifying the product $\sqrt{50} \cdot \sqrt{22}$ gives $\sqrt{1100} = 10\sqrt{11}$, which leads to $1 = 10\sin\theta$, meaning $\sin\theta = \frac{1}{10}$.

Using the identity $\cos\theta = \sqrt{1 - \sin^2\theta}$, we determine $\cos\theta = \sqrt{1 - (\frac{1}{10})^2} = \frac{\sqrt{99}}{10}$. The expression we need to evaluate, $|72 - |\vec{b} + \vec{c}|^2|$, involves the squared magnitude $|\vec{b} + \vec{c}|^2 = |\vec{b}|^2 + |\vec{c}|^2 + 2|\vec{b}||\vec{c}|\cos\theta$. Plugging in the known values, we calculate $50 + 22 + 2(\sqrt{50})(\sqrt{22})(\frac{\sqrt{99}}{10})$. This simplifies to $72 + 2(5\sqrt{2})(\sqrt{2}\sqrt{11})(\frac{3\sqrt{11}}{10})$, which further reduces to $72 + 66 = 138$. Finally, the value $|72 - 138|$ yields 66.

Since the vector $\vec{d}$ is perpendicular to both $\vec{b}$ and $\vec{c}$, it must be collinear with their cross product, allowing us to express it as $\vec{d} = \lambda (\vec{b} \times \vec{c})$ for some scalar constant $\lambda$. Computing the cross product $\vec{b} \times \vec{c}$ yields $2\hat{i} - \hat{j} + 2\hat{k}$. The provided condition $\vec{a} \cdot \vec{d} = 18$ links to the scalar triple product, such that $\vec{a} \cdot (\lambda (\vec{b} \times \vec{c})) = \lambda [\vec{a} \vec{b} \vec{c}] = 18$. Evaluating the determinant

results in $4 - 3 + 8 = 9$, which implies $9\lambda = 18$, so $\lambda = 2$ and $\vec{d} = 4\hat{i} - 2\hat{j} + 4\hat{k}$.

To calculate the squared magnitude of the cross product $|\vec{a} \times \vec{d}|^2$ efficiently, we can employ Lagrange's Identity, which states that $|\vec{a} \times \vec{d}|^2 = |\vec{a}|^2 |\vec{d}|^2 - (\vec{a} \cdot \vec{d})^2$. The squared magnitude of $\vec{a}$ is $|\vec{a}|^2 = 2^2 + 3^2 + 4^2 = 29$, and the squared magnitude of $\vec{d}$ is $|\vec{d}|^2 = 4^2 + (-2)^2 + 4^2 = 36$. Substituting these values along with the given dot product $\vec{a} \cdot \vec{d} = 18$ into the identity gives $29 \times 36 - 18^2 = 1044 - 324 = 720$.

The magnitude of a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ is defined by $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$. If we define $M = \max\{|a_1|, |a_2|, |a_3|\}$, we observe that $a_1^2 + a_2^2 + a_3^2 \geq M^2$ because the sum of squares is at least as large as the square of the largest component, which implies $|\vec{a}| \geq M$ and confirms the first statement. Regarding the second statement, since each component satisfies $a_i^2 \leq M^2$, their sum is bounded by $a_1^2 + a_2^2 + a_3^2 \leq 3M^2$, leading to $|\vec{a}| \leq \sqrt{3}M$. Because $\sqrt{3} < 3$, it is mathematically certain that $|\vec{a}| \leq 3M$, rendering the second statement true as well.

Three points are collinear if the vectors connecting them share identical direction ratios, implying their components are proportional. For the points $(\alpha, 10, 13)$, $(6, 11, 11)$, and $(\frac{9}{2}, \beta, -8)$, we determine the direction ratios of vector $\vec{AB}$ as $(6 - \alpha, 1, -2)$ and vector $\vec{BC}$ as $(-\frac{3}{2}, \beta - 11, -19)$. Setting these proportional gives the relationship:
$\frac{6 - \alpha}{-1.5} = \frac{1}{\beta - 11} = \frac{-2}{-19}$
Solving $\frac{6 - \alpha}{-1.5} = \frac{2}{19}$ yields $6 - \alpha = -\frac{3}{19}$, which simplifies to $\alpha = \frac{117}{19}$, and solving $\frac{1}{\beta - 11} = \frac{2}{19}$ results in $2\beta = 41$. Finally, substituting these into the expression $(19\alpha - 6\beta)^2$ leads to $(117 - 3(41))^2$, which is $(-6)^2 = 36$.

The direction of the normal to a plane is obtained by the cross product of two vectors lying on the plane, which serves as the foundation for identifying the orientation of the unit vector $\overset{\text{smile}}{OP}$. Using the given points $(1, 2, 3)$, $(2, 3, 4)$, and $(1, 5, 7)$, we form the vectors $\vec{u} = \langle 1, 1, 1 \rangle$ and $\vec{v} = \langle 0, 3, 4 \rangle$. Their cross product is $\vec{n} = \vec{u} \times \vec{v} = \langle 1, -4, 3 \rangle$. This vector provides the direction ratios of the normal to the plane, leading to the plane equation $1(x-1) - 4(y-2) + 3(z-3) = 0$, which simplifies to $x - 4y + 3z = 2$.

As the unit vector $\overset{\text{smile}}{OP}$ is perpendicular to the plane, it must be parallel to the normal vector $\vec{n}$, meaning it is defined by $\pm \frac{\vec{n}}{|\vec{n}|} = \pm \langle \frac{1}{\sqrt{26}}, \frac{-4}{\sqrt{26}}, \frac{3}{\sqrt{26}} \rangle$. The constraint $\beta \in \left(0, \frac{\pi}{2}\right)$ dictates that $\cos \beta$ must be positive, which is only satisfied by choosing the negative sign, resulting in $\overset{\text{smile}}{OP} = \langle \frac{-1}{\sqrt{26}}, \frac{4}{\sqrt{26}}, \frac{-3}{\sqrt{26}} \rangle$. Comparing this to the standard form $\langle \cos \alpha, \cos \beta, \cos \gamma \rangle$, we identify $\cos \alpha = \frac{-1}{\sqrt{26}}$ and $\cos \gamma = \frac{-3}{\sqrt{26}}$. Since both values are negative, the angles $\alpha$ and $\gamma$ must both reside in the interval $\left(\frac{\pi}{2}, \pi\right)$.

The area of a quadrilateral in three-dimensional space can be determined by its diagonals. For any four points $A, B, C, D$, the area is given by half the magnitude of the cross product of the vectors representing the two diagonals, $\vec{AC}$ and $\vec{BD}$.

First, we calculate the diagonal vectors by subtracting the coordinates of the endpoints: $\vec{AC} = C - A = (-4, -4, 4)$ and $\vec{BD} = D - B = (0, -8, -12)$. The cross product of these two vectors is calculated using the determinant:
$\vec{AC} \times \vec{BD} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -4 & -4 & 4 \\ 0 & -8 & -12 \end{vmatrix}$
Evaluating this determinant results in $\hat{i}(48 - (-32)) - \hat{j}(48 - 0) + \hat{k}(32 - 0)$, which simplifies to the vector $80\hat{i} - 48\hat{j} + 32\hat{k}$.

The area is half the magnitude of this vector, calculated as $\frac{1}{2} \sqrt{80^2 + (-48)^2 + 32^2}$. Simplifying the squares leads to $\frac{1}{2} \sqrt{6400 + 2304 + 1024}$, which is $\frac{1}{2} \sqrt{9728}$. Recognizing that $9728 = 256 \times 38$, the expression simplifies to $\frac{1}{2} \times 16 \sqrt{38}$, resulting in a final area of $8 \sqrt{38}$.

The relationship between the magnitudes of the cross product and the dot product of two vectors is governed by the identity $|\vec{a} \times \vec{b}|^2 + (\vec{a} \cdot \vec{b})^2 = |\vec{a}|^2 |\vec{b}|^2$. Squaring the given magnitudes, we find that $|\vec{a}|^2 = 14$, $|\vec{b}|^2 = 6$, and $|\vec{a} \times \vec{b}|^2 = 48$. Multiplying the squared magnitudes of the individual vectors yields $|\vec{a}|^2 |\vec{b}|^2 = 14 \times 6 = 84$. Subtracting the squared magnitude of the cross product from this product results in $(\vec{a} \cdot \vec{b})^2 = 84 - 48 = 36$.

Four points in space are coplanar if the scalar triple product of the vectors formed by connecting one point to the other three is equal to zero, as this condition ensures the volume of the parallelepiped they define vanishes. Using the first point as the origin for our vectors, we determine the vectors $\vec{AB} = (-2, 6, -3)$, $\vec{AC} = (-5, 3, 1)$, and $\vec{AD} = (2, 4-2\alpha, 2)$ by subtracting the position vector of the first point from the others. These points lie on the same plane if the determinant formed by these component vectors is zero:

$\begin{vmatrix} -2 & 6 & -3 \\ -5 & 3 & 1 \\ 2 & 4-2\alpha & 2 \end{vmatrix} = 0$

Expanding along the first row gives $-2(6 - (4-2\alpha)) - 6(-10 - 2) - 3(-5(4-2\alpha) - 6) = 0$, which simplifies to $-2(2 + 2\alpha) + 72 - 3(-26 + 10\alpha) = 0$. Distributing the terms leads to $-4 - 4\alpha + 72 + 78 - 30\alpha = 0$, which combines into $-34\alpha + 146 = 0$. Solving for $\alpha$ provides the final result $\alpha = \frac{73}{17}$.

\;{ {\text{A}}}^{\to }=2 { {\text{i}}^{\^}}+3 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}} \;{ {\text{B}}}^{\to }-{ {\text{A}}}^{\to }=2 { {\text{j}}^{\^}} \;{ {\text{B}}}^{\to }=2 { {\text{j}}^{\^}}+{ {\text{A}}}^{\to }=2 { {\text{i}}^{\^}}+5 { {\text{j}}^{\^}}+2 { {\text{k}}^{\^}} $\;|{ {\text{B}}}^{\to }|={\sqrt{2^{2+}5^{2+}2^2}}={\sqrt{4+25+4}}={\sqrt{8+25}}={\sqrt{33}}$

The equality of cross products $\vec{a} \times \vec{c} = \vec{b} \times \vec{c}$ indicates that the vector $(\vec{a} - \vec{b})$ is collinear with $\vec{c}$, which can be expressed as $\vec{a} - \vec{b} = \mu \vec{c}$ for some scalar $\mu$. Computing the difference yields $\vec{a} - \vec{b} = -2\hat{i} + 7\hat{j} + 2\lambda\hat{k}$. By taking the dot product of this relationship with $\vec{a}$ and $\vec{b}$ respectively, and utilizing the given values $\vec{a} \cdot \vec{c} = 7$ and $\vec{b} \cdot \vec{c} = -\frac{43}{2}$, we derive the two equations $\vec{a} \cdot (\vec{a} - \vec{b}) = 7\mu$ and $\vec{b} \cdot (\vec{a} - \vec{b}) = -\frac{43}{2}\mu$.

Substituting the identities $\vec{a} \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - \vec{a} \cdot \vec{b}$ and $\vec{b} \cdot (\vec{a} - \vec{b}) = \vec{a} \cdot \vec{b} - |\vec{b}|^2$, along with the expressions $|\vec{a}|^2 = 5 + \lambda^2$, $|\vec{b}|^2 = 34 + \lambda^2$, and $\vec{a} \cdot \vec{b} = -7 - \lambda^2$, allows us to isolate the variables. Summing the two expressions yields $|\vec{a}|^2 - |\vec{b}|^2 = (7 - \frac{43}{2})\mu$, which simplifies to $(5 + \lambda^2) - (34 + \lambda^2) = -\frac{29}{2}\mu$. This results in $-29 = -\frac{29}{2}\mu$, identifying $\mu = 2$. With $\mu$ established, substituting it back into the primary equations confirms $\lambda^2 = 1$, ultimately yielding the magnitude $|\vec{a} \cdot \vec{b}| = |-7 - 1| = 8$.

The cross product $\vec{v} \times \vec{w}$ is geometrically perpendicular to both $\vec{v}$ and $\vec{w}$, a property that allows us to eliminate terms using the dot product. Starting with the vector equation $\vec{v} \times \vec{w} = \vec{u} + \lambda\vec{v}$, taking the dot product of both sides with $\vec{w}$ makes the term $\vec{w} \cdot (\vec{v} \times \vec{w})$ vanish, leading to the relation $0 = \vec{u} \cdot \vec{w} + \lambda(\vec{v} \cdot \vec{w})$. Since $\vec{v} \cdot \vec{w}$ is given as 2, this simplifies to $\vec{u} \cdot \vec{w} = -2\lambda$.

To determine $\lambda$, apply the dot product with $\vec{v}$ to the same original equation, which forces the term $\vec{v} \cdot (\vec{v} \times \vec{w})$ to 0, resulting in $0 = \vec{u} \cdot \vec{v} + \lambda |\vec{v}|^2$. Calculating the components, the dot product $\vec{u} \cdot \vec{v}$ equals 3, and the squared magnitude $|\vec{v}|^2$ equals 6. Substituting these into the equation $0 = 3 + 6\lambda$ yields $\lambda = -1/2$. Substituting this value back into the previous relation gives $\vec{u} \cdot \vec{w} = -2(-1/2) = 1$.

The area of any triangle can be calculated using the formula $\text{Area} = \frac{1}{2}ab \sin(\theta)$, where $a$ and $b$ are the lengths of two sides forming an angle $\theta$. When a new triangle is inscribed within a larger one, the area of the inner triangle can be determined by subtracting the areas of the three corner triangles from the total area. The ratio of the area of each corner triangle to the large triangle is equal to the product of the ratios of the lengths of the two sides framing the corner relative to the original sides.

Starting with vertex $Q$, the corner triangle $\triangle CQA$ has sides $QC$ and $QA$ adjacent to the angle at $Q$. Given the ratios $\frac{PC}{CQ} = \frac{1}{2}$ and $\frac{QA}{AR} = \frac{1}{2}$, we find that $QC = \frac{2}{3}PQ$ and $QA = \frac{1}{3}QR$. Thus, the area of $\triangle CQA$ is $\left(\frac{2}{3}\right) \left(\frac{1}{3}\right) = \frac{2}{9}$ of the total area of $\triangle PQR$. Similarly, for vertex $R$, the triangle $\triangle ARB$ uses segments $AR = \frac{2}{3}QR$ and $RB = \frac{1}{3}RP$, making its area $\frac{2}{9}$ of the total. At vertex $P$, the triangle $\triangle BPC$ uses sides $PB = \frac{2}{3}RP$ and $PC = \frac{1}{3}PQ$, also yielding an area that is $\frac{2}{9}$ of the whole.

Combining these three corner areas gives $\frac{2}{9} + \frac{2}{9} + \frac{2}{9} = \frac{6}{9} = \frac{2}{3}$ of the total area. The area of the remaining inner triangle $\triangle ABC$ is therefore $\left(1 - \frac{2}{3}\right) = \frac{1}{3}$ of the total area of $\triangle PQR$. Consequently, the ratio $\frac{\text{Area}(\triangle PQR)}{\text{Area}(\triangle ABC)}$ is equal to $3$.

\vec{\text{P}} \times \vec{\text{Q}}={\begin{vmatrix}{\text{i}}^{\^} & {\text{j}}^{\^} & {\text{k}}^{\^} \\ 3 & {\sqrt{3}} & 2 \\ 4 & {\sqrt{3}} & 2.5\end{vmatrix}} ={\sqrt{3}} \;{{ {\text{i}}^{\^}}}/{2}+\;{{ {\text{j}}^{\^}}}/{2}-{\sqrt{3}} { {\text{k}}^{\^}} \Rightarrow \;{{ {\text{P}}}^{\to } \times { {\text{Q}}}^{\to }}/{|{ {\text{P}}}^{\to } \times { {\text{Q}}}^{\to }|}=\;\frac{1}{2} ({\sqrt{3}} \;{{ {\text{i}}^{\^}}}/{2}+\;{{ {\text{j}}^{\^}}}/{2}-{\sqrt{3}} { {\text{k}}^{\^}}) \;=\;\frac{1}{4}({\sqrt{3}} { {\text{i}}^{\^}+{ {\text{j}}^{\^}-2 {\sqrt{3}} { {\text{k}}^{\^}) \;\; {\text{x}}=4

Vector geometry problems involving arbitrary quadrilaterals are best approached by representing vertices with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ and identifying the midpoints as $\vec{e} = \frac{\vec{a} + \vec{c}}{2}$ and $\vec{f} = \frac{\vec{b} + \vec{d}}{2}$. Expanding the expression $(\vec{AB} - \vec{BC}) + (\vec{AD} - \vec{DC})$ in terms of these vectors simplifies the sum to $2\vec{b} + 2\vec{d} - 2\vec{a} - 2\vec{c}$, which can be rewritten as $4\vec{f} - 4\vec{e}$. Recognizing that $\vec{FE} = \vec{e} - \vec{f}$ allows us to factor out the negative sign to obtain $-4(\vec{e} - \vec{f})$, which is equal to $-4\vec{FE}$. Thus, the constant $k$ is $-4$.

Three vectors are coplanar if and only if their scalar triple product is zero, which is equivalent to saying that the determinant of the matrix formed by these vectors as rows must vanish. Calculating this determinant for the vectors $\lambda \hat{i} - \hat{j} + \hat{k}$, $\hat{i} + 2 \hat{j} + \mu \hat{k}$, and $3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ leads to the equation

. Expanding this determinant along the first row yields $\lambda(10 + 4\mu) + 1(5 - 3\mu) + 1(-4 - 6) = 0$, which simplifies to $10\lambda + 4\lambda\mu + 5 - 3\mu - 10 = 0$, or $10\lambda + 4\lambda\mu - 3\mu - 5 = 0$.

Given the constraint $\lambda - \mu = 5$, we substitute $\lambda = \mu + 5$ into this expression to solve for the permissible values of the variables. This substitution results in $10(\mu + 5) + 4\mu(\mu + 5) - 3\mu - 5 = 0$, which further simplifies to the quadratic equation $4\mu^2 + 27\mu + 45 = 0$. Factoring this quadratic gives $(4\mu + 15)(\mu + 3) = 0$, providing the solutions $\mu = -3$ and $\mu = -15/4$. Determining the corresponding values for $\lambda$ using $\lambda = \mu + 5$, we find the pairs $(\lambda, \mu)$ to be $(2, -3)$ and $(5/4, -15/4)$.

The problem asks for the sum $80(\lambda^2 + \mu^2)$ for both pairs. For the first pair, $80(2^2 + (-3)^2) = 80(4 + 9) = 80(13) = 1040$. For the second pair, $80((5/4)^2 + (-15/4)^2) = 80(\frac{25}{16} + \frac{225}{16}) = 80(\frac{250}{16}) = 5 \times 250 = 1250$. Adding these two results together gives a final value of $1040 + 1250 = 2290$.

The volume of a parallelepiped defined by three coterminous edges is determined by the absolute value of the scalar triple product of those vectors. When the original edges $\vec{a}, \vec{b}, \vec{c}$ are replaced by new vectors, the ratio of the new volume to the original volume $V$ is equal to the determinant of the transformation matrix that maps the original basis to the new edge vectors. In this problem, the new edges are $\vec{a}$, $\vec{b}+\vec{c}$, and $\vec{a}+2\vec{b}+3\vec{c}$, which can be represented by the coefficient matrix

. Calculating this determinant yields $1(3-2) - 0(0) + 0(0) = 1$, which indicates that the magnitude of the scalar triple product remains unchanged. Therefore, the volume of the new parallelepiped is equal to $V$.

\;{ \vec{\text{P}} \cdot { \vec{\text{Q}}=0 \;(\;{ {\text{i}}}^{\^}+2 {\text{m}} \;{ {\text{j}}}^{\^}+ {\text{m}} \;{ {\text{k}}}^{\^}) \cdot (4 \;{ {\text{i}}}^{\^}-2 \;{ {\text{j}}}^{\^}+ {\text{m}} \;{ {\text{k}}}^{\^})=0 $\;\Rightarrow 4-4 {\text{m}}+ {\text{m}}^2=0$ $\;\Rightarrow ( {\text{m}}-2)^2=0 \Rightarrow {\text{m}}=2$