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Find the value of m for which the vectors -2mi + 2j - 2k and 3i + 4mj + 3k perpendicular to each other.

Mathematics

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Step-by-step solution

11 steps
  1. Identify the two vectors.

    Let A⃗=−2mi⃗+2j⃗−2k⃗\vec{A} = -2m\vec{i} + 2\vec{j} - 2\vec{k} and B⃗=3i⃗+4mj⃗+3k⃗\vec{B} = 3\vec{i} + 4m\vec{j} + 3\vec{k}.

  2. Apply the condition for perpendicular vectors.

    Two vectors are perpendicular if their dot product is zero (A⃗⋅B⃗=0\vec{A} \cdot \vec{B} = 0).

    (−2m)(3)+(2)(4m)+(−2)(3)=0(-2m)(3) + (2)(4m) + (-2)(3) = 0
  3. Simplify the equation.

    −6m+8m−6=0-6m + 8m - 6 = 0 2m−6=02m - 6 = 0
  4. Solve for mm.

    2m=62m = 6 m=62m = \frac{6}{2} m=3m = 3

    The value of mm is 3\boxed{3}.


  5. Find the resultant vector R⃗\vec{R}.

    The resultant vector is the sum of a⃗1\vec{a}_1 and a⃗2\vec{a}_2.

    R⃗=a⃗1+a⃗2=(6i⃗−3j⃗+2k⃗)+(3i⃗+j⃗−2k⃗)\vec{R} = \vec{a}_1 + \vec{a}_2 = (6\vec{i} - 3\vec{j} + 2\vec{k}) + (3\vec{i} + \vec{j} - 2\vec{k}) R⃗=(6+3)i⃗+(−3+1)j⃗+(2−2)k⃗\vec{R} = (6+3)\vec{i} + (-3+1)\vec{j} + (2-2)\vec{k} R⃗=9i⃗−2j⃗+0k⃗\vec{R} = 9\vec{i} - 2\vec{j} + 0\vec{k}
  6. Calculate the magnitude of the resultant vector ∣R⃗∣|\vec{R}|.

    ∣R⃗∣=92+(−2)2+02|\vec{R}| = \sqrt{9^2 + (-2)^2 + 0^2} ∣R⃗∣=81+4+0|\vec{R}| = \sqrt{81 + 4 + 0} ∣R⃗∣=85|\vec{R}| = \sqrt{85}
  7. Calculate the unit vector parallel to R⃗\vec{R}.

    The unit vector u^\hat{u} is R⃗∣R⃗∣\frac{\vec{R}}{|\vec{R}|}.

    u^=9i⃗−2j⃗85\hat{u} = \frac{9\vec{i} - 2\vec{j}}{\sqrt{85}} u^=985i⃗−285j⃗\hat{u} = \frac{9}{\sqrt{85}}\vec{i} - \frac{2}{\sqrt{85}}\vec{j}

    The unit vector parallel to the resultant is ∗985i⃗−285j⃗∗\boxed*{\frac{9}{\sqrt{85}}\vec{i} - \frac{2}{\sqrt{85}}\vec{j}*}.


  8. Identify the two vectors.

    Let A⃗=3i⃗−4j⃗+12k⃗\vec{A} = 3\vec{i} - 4\vec{j} + 12\vec{k} and B⃗=i⃗+2j⃗−k⃗\vec{B} = \vec{i} + 2\vec{j} - \vec{k}.

  9. Calculate the cross product C⃗=A⃗×B⃗\vec{C} = \vec{A} \times \vec{B}.

    C⃗=∣i⃗j⃗k⃗3−41212−1∣\vec{C} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 3 & -4 & 12 \\ 1 & 2 & -1 \end{vmatrix} C⃗=i⃗((−4)(−1)−(12)(2))−j⃗((3)(−1)−(12)(1))+k⃗((3)(2)−(−4)(1))\vec{C} = \vec{i}((-4)(-1) - (12)(2)) - \vec{j}((3)(-1) - (12)(1)) + \vec{k}((3)(2) - (-4)(1)) C⃗=i⃗(4−24)−j⃗(−3−12)+k⃗(6+4)\vec{C} = \vec{i}(4 - 24) - \vec{j}(-3 - 12) + \vec{k}(6 + 4) C⃗=−20i⃗+15j⃗+10k⃗\vec{C} = -20\vec{i} + 15\vec{j} + 10\vec{k}
  10. Calculate the magnitude of C⃗\vec{C}.

    ∣C⃗∣=(−20)2+152+102|\vec{C}| = \sqrt{(-20)^2 + 15^2 + 10^2} ∣C⃗∣=400+225+100|\vec{C}| = \sqrt{400 + 225 + 100} ∣C⃗∣=725|\vec{C}| = \sqrt{725} ∣C⃗∣=25×29=529|\vec{C}| = \sqrt{25 \times 29} = 5\sqrt{29}
  11. Calculate the unit vector perpendicular to A⃗\vec{A} and B⃗\vec{B}.

    The unit vector u^\hat{u} is C⃗∣C⃗∣\frac{\vec{C}}{|\vec{C}|}.

    u^=−20i⃗+15j⃗+10k⃗529\hat{u} = \frac{-20\vec{i} + 15\vec{j} + 10\vec{k}}{5\sqrt{29}} u^=−20529i⃗+15529j⃗+10529k⃗\hat{u} = \frac{-20}{5\sqrt{29}}\vec{i} + \frac{15}{5\sqrt{29}}\vec{j} + \frac{10}{5\sqrt{29}}\vec{k} u^=−429i⃗+329j⃗+229k⃗\hat{u} = -\frac{4}{\sqrt{29}}\vec{i} + \frac{3}{\sqrt{29}}\vec{j} + \frac{2}{\sqrt{29}}\vec{k}

    The unit vector perpendicular to the given vectors is ∗−429i⃗+329j⃗+229k⃗∗\boxed*{-\frac{4}{\sqrt{29}}\vec{i} + \frac{3}{\sqrt{29}}\vec{j} + \frac{2}{\sqrt{29}}\vec{k}*}.


    i. Magnitude of a vector
    The magnitude of a vector is its length or size, representing the scalar value of the vector quantity. For a vector v⃗=xi⃗+yj⃗+zk⃗\vec{v} = x\vec{i} + y\vec{j} + z\vec{k}, its magnitude is given by ∣v⃗∣=x2+y2+z2|\vec{v}| = \sqrt{x^2 + y^2 + z^2}.

    ii. Directional Cosines
    Directional cosines are the cosines of the angles that a vector makes with the positive x, y, and z axes. For a vector v⃗=xi⃗+yj⃗+zk⃗\vec{v} = x\vec{i} + y\vec{j} + z\vec{k}, the directional cosines are cos⁡α=x∣v⃗∣\cos\alpha = \frac{x}{|\vec{v}|}, cos⁡β=y∣v⃗∣\cos\beta = \frac{y}{|\vec{v}|}, and cos⁡γ=z∣v⃗∣\cos\gamma = \frac{z}{|\vec{v}|}.

    iii. Parallelism of two vectors
    Two vectors are parallel if they point in the same or opposite direction. Mathematically, two non-zero vectors A⃗\vec{A} and B⃗\vec{B} are parallel if one is a scalar multiple of the other, i.e., A⃗=kB⃗\vec{A} = k\vec{B} for some scalar kk.

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