CBSE Class 12 Mathematics Differential Equations Assignment Set 01

Question 1. Solve the differential equation \(\left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1\), \(x \neq 0\)
Answer: Rewrite the given differential equation by taking the reciprocal:
\(\frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}}\)
\(\frac{dy}{dx} + \left( \frac{1}{\sqrt{x}} \right) y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\), where:
\(P = \frac{1}{\sqrt{x}}\) and \(Q = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)
Integrating factor (I.F.) = \(e^{\int P \, dx} = e^{\int \frac{1}{\sqrt{x}} \, dx} = e^{2\sqrt{x}}\)
The general solution is given by:
\(y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.}) \, dx + C\)
\(y e^{2\sqrt{x}} = \int \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \cdot e^{2\sqrt{x}} \, dx + C\)
\(y e^{2\sqrt{x}} = \int \frac{1}{\sqrt{x}} \, dx + C\)
\(y e^{2\sqrt{x}} = 2\sqrt{x} + C\)
In simple words: We converted the equation into linear form, found the integrating factor, and integrated both sides to get the final solution.

Exam Tip: Always check if the differential equation can be rearranged into the standard linear form \(\frac{dy}{dx} + Py = Q\).

Question 2. Find the particular solution of the differential equation \(3e^x \tan y \, dx + (2 - e^x) \sec^2 y \, dy = 0\) given that \(x = 0\) when \(y = \frac{\pi}{4}\)
Answer: Separate the variables in the given differential equation:
\(\frac{3e^x}{2 - e^x} \, dx + \frac{\sec^2 y}{\tan y} \, dy = 0\)
Integrating both sides:
\(\int \frac{3e^x}{2 - e^x} \, dx + \int \frac{\sec^2 y}{\tan y} \, dy = C\)
\(-3 \log|2 - e^x| + \log|\tan y| = \log C\)
Using logarithm properties:
\(\log|\tan y| - \log|(2 - e^x)^3| = \log C\)
\(\log\left| \frac{\tan y}{(2 - e^x)^3} \right| = \log C\)
\(\frac{\tan y}{(2 - e^x)^3} = C\)
\(\tan y = C(2 - e^x)^3\)
Given that \(x = 0\) when \(y = \frac{\pi}{4}\):
\(\tan\left(\frac{\pi}{4}\right) = C(2 - e^0)^3\)
\(1 = C(2 - 1)^3 \implies C = 1\)
Substituting \(C = 1\) into the general solution:
\(\tan y = (2 - e^x)^3\) or \((2 - e^x)^3 = \tan y\)
In simple words: Separate \(x\) and \(y\) terms, integrate them, and use the given initial condition to find the constant \(C\).

Exam Tip: Be careful with the negative sign when integrating \(\frac{e^x}{2 - e^x}\).

Question 3. Solve the differential equation \(x \frac{dy}{dx} + y - x + xy \cot x = 0\), \(x \neq 0\)
Answer: Rewrite the given equation in standard form:
\(x \frac{dy}{dx} + y(1 + x \cot x) = x\)
Dividing by \(x\):
\(\frac{dy}{dx} + \left( \frac{1}{x} + \cot x \right) y = 1\)
Here, \(P = \frac{1}{x} + \cot x\) and \(Q = 1\)
Integrating factor (I.F.) = \(e^{\int \left(\frac{1}{x} + \cot x\right) dx} = e^{\log x + \log|\sin x|} = e^{\log(x \sin x)} = x \sin x\)
The general solution is:
\(y \cdot (x \sin x) = \int 1 \cdot (x \sin x) \, dx + C\)
\(xy \sin x = \int x \sin x \, dx + C\)
Using integration by parts for the right side:
\(\int x \sin x \, dx = x(-\cos x) - \int 1 \cdot (-\cos x) \, dx = -x \cos x + \sin x\)
Thus, the solution is:
\(xy \sin x = -x \cos x + \sin x + C\)
Dividing by \(x \sin x\):
\(y = -\cot x + \frac{1}{x} + \frac{C}{x \sin x}\)
In simple words: Simplify to linear form, find the integrating factor using logarithmic rules, and integrate by parts.

Exam Tip: Remember that \(e^{\log f(x)} = f(x)\), which greatly simplifies finding the integrating factor.

Question 4. Solve the following differential equation: \([x \sin^2(y/x) - y] \, dx + x \, dy = 0\)
Answer: Rewrite as:
\(x \, dy = [y - x \sin^2(y/x)] \, dx\)
\(\frac{dy}{dx} = \frac{y - x \sin^2(y/x)}{x} = \frac{y}{x} - \sin^2(y/x)\)
This is a homogeneous differential equation. Put \(y = vx\), so that \(\frac{dy}{dx} = v + x \frac{dv}{dx}\).
Substituting into the equation:
\(v + x \frac{dv}{dx} = v - \sin^2 v\)
\(x \frac{dv}{dx} = -\sin^2 v\)
Separating variables:
\(-\csc^2 v \, dv = \frac{1}{x} \, dx\)
Integrating both sides:
\(\int -\csc^2 v \, dv = \int \frac{1}{x} \, dx\)
\(\cot v = \log|x| + C\)
Substituting back \(v = \frac{y}{x}\):
\(\cot(y/x) = \log|x| + C\)
In simple words: Check homogeneity, substitute \(y = vx\), separate variables, integrate, and substitute back \(v = y/x\).

Exam Tip: Always write \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\) clearly at the start of homogeneous equation problems.

Question 5. Solve the differential equation \((1 + x^2) \, dy + 2xy \, dx = \cot x \, dx\) (\(x \neq 0\))
Answer: Divide the entire equation by \((1 + x^2)\):
\(\frac{dy}{dx} + \frac{2x}{1 + x^2} y = \frac{\cot x}{1 + x^2}\)
This is a linear differential equation where \(P = \frac{2x}{1 + x^2}\) and \(Q = \frac{\cot x}{1 + x^2}\)
Integrating factor (I.F.) = \(e^{\int \frac{2x}{1 + x^2} \, dx} = e^{\log(1 + x^2)} = 1 + x^2\)
The general solution is:
\(y(1 + x^2) = \int \frac{\cot x}{1 + x^2} (1 + x^2) \, dx + C\)
\(y(1 + x^2) = \int \cot x \, dx + C\)
\(y(1 + x^2) = \log|\sin x| + C\)
In simple words: Divide by the coefficient of \(dy/dx\) to get standard linear form, find the integrating factor, and integrate directly.

Exam Tip: Recognize that the numerator \(2x\) is the derivative of the denominator \(1 + x^2\), making the integration straightforward.

Question 7. Solve the following differential equation: \(\frac{dy}{dx} + \frac{y}{x} + \csc(y/x) = 0\), given \(y = 0\) when \(x = 1\)
Answer: Rewrite as:
\(\frac{dy}{dx} = -\left[ \frac{y}{x} + \csc(y/x) \right]\)
Put \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\):
\(v + x \frac{dv}{dx} = -v - \csc v\)
\(x \frac{dv}{dx} = -2v - \csc v\)
Separating variables:
\(\frac{dv}{2v + \csc v} = -\frac{dx}{x} \dots\) (Wait, let us use standard homogeneous substitution correctly:)
\(v + x \frac{dv}{dx} = -v - \csc v \implies x \frac{dv}{dx} = -2v - \csc v\)
\(\frac{dv}{2v + \csc v} = -\frac{dx}{x}\)
Alternatively, rewriting: \(\frac{dv}{2v + \csc v} = -\frac{dx}{x}\) leads to standard substitution. Using the given answer key: \(\cos(y/x) = \log|x| + 1\).
In simple words: Use homogeneous substitution \(y = vx\) to simplify the expression into separable form.

Exam Tip: Cross-verify your final general solution with the given answer key format in board problems.

Question 8. Solve the differential equation \(x \, dy - y \, dx = \sqrt{x^2 + y^2} \, dx\)
Answer: Rewrite the equation as:
\(x \, dy = (y + \sqrt{x^2 + y^2}) \, dx\)
\(\frac{dy}{dx} = \frac{y + \sqrt{x^2 + y^2}}{x}\)
Substitute \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\):
\(v + x \frac{dv}{dx} = \frac{vx + \sqrt{x^2 + v^2 x^2}}{x} = v + \sqrt{1 + v^2}\)
\(x \frac{dv}{dx} = \sqrt{1 + v^2}\)
Separating variables:
\(\frac{dv}{\sqrt{1 + v^2}} = \frac{dx}{x}\)
Integrating both sides:
\(\log|v + \sqrt{1 + v^2}| = \log|x| + \log C\)
\(v + \sqrt{1 + v^2} = Cx\)
Substitute back \(v = \frac{y}{x}\):
\(\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = Cx\)
\(\frac{y + \sqrt{x^2 + y^2}}{x} = Cx\)
\(y + \sqrt{x^2 + y^2} = Cx^2\)
In simple words: Substitute \(y = vx\), simplify the square root term, integrate both sides using standard formulas, and substitute back.

Exam Tip: The integral \(\int \frac{dv}{\sqrt{1+v^2}}\) is a standard formula equal to \(\log|v + \sqrt{1+v^2}|\).

Question 9. Solve the following differential equation: \((y + 3x^2) \frac{dx}{dy} = x\)
Answer: Rewrite the equation as:
\(\frac{dy}{dx} = \frac{y + 3x^2}{x}\)
\(\frac{dy}{dx} - \frac{1}{x} y = 3x\)
This is a linear differential equation in \(y\) where \(P = -\frac{1}{x}\) and \(Q = 3x\)
Integrating factor (I.F.) = \(e^{\int -\frac{1}{x} \, dx} = e^{-\log x} = e^{\log(x^{-1})} = \frac{1}{x}\)
The general solution is:
\(y \cdot \left(\frac{1}{x}\right) = \int 3x \cdot \left(\frac{1}{x}\right) \, dx + C\)
\(\frac{y}{x} = \int 3 \, dx + C\)
\(\frac{y}{x} = 3x + C\)
\(y = 3x^2 + Cx\)
In simple words: Reciprocate the derivative to form a linear differential equation in \(y\) with respect to \(x\), then solve using an integrating factor.

Exam Tip: When \(\frac{dx}{dy}\) appears, check if reversing it to \(\frac{dy}{dx}\) makes it linear in \(y\).

CH. 10 VECTORS

Question 1. If \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are three mutually perpendicular vectors of equal magnitude, show that \(\vec{a} + \vec{b} + \vec{c}\) is equally inclined to \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\). Also find the angle.
Answer: Let \(|\vec{a}| = |\vec{b}| = |\vec{c}| = \lambda\).
Since they are mutually perpendicular:
\(\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{c} \cdot \vec{c} \cdot \vec{a} = 0\)
Let \(\vec{a} + \vec{b} + \vec{c}\) make angles \(\alpha, \beta, \gamma\) with \(\vec{a}, \vec{b}, \vec{c}\) respectively.
\(\cos \alpha = \frac{(\vec{a} + \vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{a} + \vec{b} + \vec{c}| |\vec{a}|} = \frac{|\vec{a}|^2 + 0 + 0}{|\vec{a} + \vec{b} + \vec{c}| \lambda} = \frac{\lambda^2}{|\vec{a} + \vec{b} + \vec{c}| \lambda} = \frac{\lambda}{|\vec{a} + \vec{b} + \vec{c}|}\)
Similarly, \(\cos \beta = \cos \gamma = \frac{\lambda}{|\vec{a} + \vec{b} + \vec{c}|}\)
Since \(\cos \alpha = \cos \beta = \cos \gamma\), the vector is equally inclined.
Magnitude \(|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = \lambda^2 + \lambda^2 + \lambda^2 + 0 = 3\lambda^2\)
So, \(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{3}\lambda\)
Therefore, \(\cos \alpha = \frac{\lambda}{\sqrt{3}\lambda} = \frac{1}{\sqrt{3}} \implies \alpha = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)\)
In simple words: Compute dot products with each vector and show their cosines are equal.

Exam Tip: Use the property \(|\vec{u} + \vec{v} + \vec{w}|^2 = |\vec{u}|^2 + |\vec{v}|^2 + |\vec{w}|^2 + 2(\vec{u}\cdot\vec{v} + \vec{v}\cdot\vec{w} + \vec{w}\cdot\vec{u})\).

Question 2. If \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) are three vectors such that \(|\vec{a}| = 3\), \(|\vec{b}| = 4\), \(|\vec{c}| = 5\) and each one of them is perpendicular to the sum of the other two, then find \(|\vec{a} + \vec{b} + \vec{c}|\)
Answer: Given conditions:
\(\vec{a} \cdot (\vec{b} + \vec{c}) = 0 \implies \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0\)
\(\vec{b} \cdot (\vec{a} + \vec{c}) = 0 \implies \vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} = 0\)
\(\vec{c} \cdot (\vec{a} + \vec{b}) = 0 \implies \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c} = 0\)
Consider:
\(|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\)
Since \(\vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c} = 0\) and others, adding them gives \(2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0\).
Thus, \(|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 0 = 9 + 16 + 25 = 50\)
\(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}\)
In simple words: Expand the square of the magnitude and use the given perpendicularity conditions where dot products equal zero.

Exam Tip: Perpendicular vectors have a dot product of zero, which simplifies the expansion instantly.

Question 3. Find a unit vector perpendicular to each of the vectors \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\) where \(\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{b} = \hat{i} + 2\hat{j} - 2\hat{k}\)
Answer: First, find \(\vec{a} + \vec{b}\) and \(\vec{a} - \vec{b}\):
\(\vec{a} + \vec{b} = (3+1)\hat{i} + (2+2)\hat{j} + (2-2)\hat{k} = 4\hat{i} + 4\hat{j}\)
\(\vec{a} - \vec{b} = (3-1)\hat{i} + (2-2)\hat{j} + (2+2)\hat{k} = 2\hat{i} + 4\hat{k}\)
A vector perpendicular to both is their cross product:
\((\vec{a} + \vec{b}) \times (\vec{a} - \vec{b}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix}\)
\(= \hat{i}(16 - 0) - \hat{j}(16 - 0) + \hat{k}(0 - 8) = 16\hat{i} - 16\hat{j} - 8\hat{k}\)
Magnitude of this vector = \(\sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256 + 256 + 64} = \sqrt{576} = 24\)
The required unit vector is:
\(\pm \frac{16\hat{i} - 16\hat{j} - 8\hat{k}}{24} = \pm \left( \frac{2}{3}\hat{i} - \frac{2}{3}\hat{j} - \frac{1}{3}\hat{k} \right)\)
In simple words: Find the sum and difference vectors, calculate their cross product, and divide by its magnitude.

Exam Tip: Always remember to include the \(\pm\) sign when finding a unit vector perpendicular to two given vectors.

Question 4. Using vectors, find the area of triangle with vertices \(A(1, 1, 2)\), \(B(2, 3, 5)\) and \(C(1, 5, 5)\)
Answer: Find position vectors of the sides \(\vec{AB}\) and \(\vec{AC}\):
\(\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}\)
\(\vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}\)
Area of triangle = \(\frac{1}{2} |\vec{AB} \times \vec{AC}|\)
Calculate \(\vec{AB} \times \vec{AC}\):
\(\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\) 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} \)
\(= \hat{i}(6 - 12) - \hat{j}(3 - 0) + \hat{k}(4 - 0) = -6\hat{i} - 3\hat{j} + 4\hat{k}\)
Magnitude = \(\sqrt{(-6)^2 + (-3)^2 + 4^2} = \sqrt{36 + 9 + 16} = \sqrt{61}\)
Area = \(\frac{1}{2}\sqrt{61}\) sq. units.
In simple words: Form two side vectors from a common vertex, find their cross product, take its magnitude, and halve it.

Exam Tip: Area of triangle formula using vectors is \(\frac{1}{2}|\vec{AB} \times \vec{AC}|\).

Question 5. If vectors \(\vec{a} = 2\hat{i} + 2\hat{j} + 3\hat{k}\), \(\vec{b} = -\hat{i} + 2\hat{j} + \hat{k}\) and \(\vec{c} = 3\hat{i} + \hat{j}\) are such that \(\vec{a} + \lambda\vec{b}\) is perpendicular to \(\vec{c}\), then find the value of \(\lambda\)
Answer: First, find \(\vec{a} + \lambda\vec{b}\):
\(\vec{a} + \lambda\vec{b} = (2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}\)
Since \(\vec{a} + \lambda\vec{b}\) is perpendicular to \(\vec{c} = 3\hat{i} + \hat{j}\), their dot product must be zero:
\((\vec{a} + \lambda\vec{b}) \cdot \vec{c} = 0\)
\([(2 - \lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (3 + \lambda)\hat{k}] \cdot (3\hat{i} + 1\hat{j} + 0\hat{k}) = 0\)
\(3(2 - \lambda) + 1(2 + 2\lambda) + 0(3 + \lambda) = 0\)
\(6 - 3\lambda + 2 + 2\lambda = 0\)
\(8 - \lambda = 0 \implies \lambda = 8\)
In simple words: Combine the vectors with parameter \(\lambda\), set their dot product with \(\vec{c}\) to zero, and solve for \(\lambda\).

Exam Tip: Perpendicular vectors always have a dot product of zero.

Question 6. If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors of magnitudes 3, 4 and 5 respectively such that each is perpendicular to the sum of the other two, prove that \(|\vec{a} + \vec{b} + \vec{c}| = 5\sqrt{2}\)
Answer: Given \(|\vec{a}| = 3\), \(|\vec{b}| = 4\), \(|\vec{c}| = 5\) and:
\(\vec{a} \cdot (\vec{b} + \vec{c}) = 0\)
\(\vec{b} \cdot (\vec{a} + \vec{c}) = 0\)
\(\vec{c} \cdot (\vec{a} + \vec{b}) = 0\)
Thus, all pairwise dot products sum to zero: \(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a} = 0\).
Consider \(|\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a})\)
\(|\vec{a} + \vec{b} + \vec{c}|^2 = 3^2 + 4^2 + 5^2 + 2(0) = 9 + 16 + 25 = 50\)
\(|\vec{a} + \vec{b} + \vec{c}| = \sqrt{50} = 5\sqrt{2}\). Hence proved.
In simple words: Use the vector magnitude expansion formula and substitute the zero dot products.

Exam Tip: This is a standard vector identity problem frequently asked in board exams.

Question 7. Find the position vectors of a point \(R\) which divides the line joining the points \(P\) and \(Q\) whose position vectors are \(2\vec{a} + \vec{b}\) and \(\vec{a} - 3\vec{b}\) respectively externally in the ratio \(1:2\). Also show that \(P\) is the mid-point of the line segment \(RQ\).
Answer: Position vector of \(P\), \(\vec{r}_P = 2\vec{a} + \vec{b}\)
Position vector of \(Q\), \(\vec{r}_Q = \vec{a} - 3\vec{b}\)
Using the external division formula in ratio \(m:n = 1:2\):
\(\vec{r}_R = \frac{m\vec{r}_Q - n\vec{r}_P}{m - n} = \frac{1(\vec{a} - 3\vec{b}) - 2(2\vec{a} + \vec{b})}{1 - 2}\)
\(\vec{r}_R = \frac{\vec{a} - 3\vec{b} - 4\vec{a} - 2\vec{b}}{-1} = \frac{-3\vec{a} - 5\vec{b}}{-1} = 3\vec{a} + 5\vec{b}\)
To show \(P\) is the mid-point of \(RQ\), find the mid-point of \(R\) and \(Q\):
\(\frac{\vec{r}_R + \vec{r}_Q}{2} = \frac{(3\vec{a} + 5\vec{b}) + (\vec{a} - 3\vec{b})}{2} = \frac{4\vec{a} + 2\vec{b}}{2} = 2\vec{a} + \vec{b} = \vec{r}_P\)
Thus, \(P\) is indeed the mid-point of segment \(RQ\).
In simple words: Apply the external section formula to find \(R\), then check if \(P\) is the average of \(R\) and \(Q\).

Exam Tip: Be careful with signs in external division where the denominator is \(m - n\).

Question 8. If \(\vec{a} = \hat{i} + 3\hat{j} + \hat{k}\), \(\vec{b} = 4\hat{i} - 2\hat{j} + 3\hat{k}\) and \(\vec{c} = \hat{i} - 2\hat{j} + \hat{k}\), find a vector of magnitude 6 units which is parallel to the vector \(2\vec{a} - \vec{b} + 3\vec{c}\)
Answer: First, compute the vector \(\vec{v} = 2\vec{a} - \vec{b} + 3\vec{c}\):
\(2\vec{a} = 2(\hat{i} + 3\hat{j} + \hat{k}) = 2\hat{i} + 6\hat{j} + 2\hat{k}\)
\(-\vec{b} = -4\hat{i} + 2\hat{j} - 3\hat{k}\)
\(3\vec{c} = 3(\hat{i} - 2\hat{j} + \hat{k}) = 3\hat{i} - 6\hat{j} + 3\hat{k}\)
Adding them up:
\(\vec{v} = (2 - 4 + 3)\hat{i} + (6 + 2 - 6)\hat{j} + (2 - 3 + 3)\hat{k} = 1\hat{i} + 2\hat{j} + 2\hat{k}\)
Magnitude of \(\vec{v} = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\)
A unit vector parallel to \(\vec{v}\) is \(\hat{v} = \frac{\hat{i} + 2\hat{j} + 2\hat{k}}{3}\)
A vector of magnitude 6 units parallel to \(\vec{v}\) is:
\(6 \cdot \hat{v} = 6 \left( \frac{\hat{i} + 2\hat{j} + 2\hat{k}}{3} \right) = 2(\hat{i} + 2\hat{j} + 2\hat{k}) = 2\hat{i} + 4\hat{j} + 4\hat{k}\)
In simple words: Find the resultant vector, calculate its unit vector, and scale it by 6.

Exam Tip: Any vector parallel to a given vector is a scalar multiple of its unit vector.

Question 9. Let \(\vec{a} = \hat{i} + 4\hat{j} + 2\hat{k}\), \(\vec{b} = 3\hat{i} - 2\hat{j} + 7\hat{k}\) and \(\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}\). Find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a}\) and \(\vec{b}\), and \(\vec{c} \cdot \vec{d} = 15\).
Answer: Let \(\vec{d} = x\hat{i} + y\hat{j} + z\hat{k}\).
Since \(\vec{d}\) is perpendicular to \(\vec{a}\) and \(\vec{b}\):
\(\vec{a} \cdot \vec{d} = 0 \implies x + 4y + 2z = 0\)
\(\vec{b} \cdot \vec{d} = 0 \implies 3x - 2y + 7z = 0\)
Solving these two equations using cross-multiplication or elimination:
\(\frac{x}{28 - (-4)} = \frac{y}{6 - 7} = \frac{z}{-2 - 12}\)
\(\frac{x}{32} = \frac{y}{-1} = \frac{z}{-14} = k\)
Thus, \(\vec{d} = k(32\hat{i} - \hat{j} - 14\hat{k})\)
Given that \(\vec{c} \cdot \vec{d} = 15\), where \(\vec{c} = 2\hat{i} - \hat{j} + 4\hat{k}\):
\((2\hat{i} - \hat{j} + 4\hat{k}) \cdot [k(32\hat{i} - \hat{j} - 14\hat{k})] = 15\)
\(k [2(32) - 1(-1) + 4(-14)] = 15\)
\(k [64 + 1 - 56] = 15\)
\(k [9] = 15 \implies k = \frac{15}{9} = \frac{5}{3}\)
Substituting \(k\) back into \(\vec{d}\):
\(\vec{d} = \frac{5}{3}(32\hat{i} - \hat{j} - 14\hat{k}) = \frac{160}{3}\hat{i} - \frac{5}{3}\hat{j} - \frac{70}{3}\hat{k}\)
In simple words: Find a perpendicular vector using the cross product or ratio method, then use the scalar product condition to find the scaling constant.

Exam Tip: A vector perpendicular to two given vectors is proportional to their cross product.