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Vector Algebra - Vector joining two points

Grade 12CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Consider two points P(x1,y1,z1)P(x_1, y_1, z_1) and Q(x2,y2,z2)Q(x_2, y_2, z_2) in a three-dimensional Cartesian coordinate system.

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The position vectors of points PP and QQ with respect to the origin O(0,0,0)O(0,0,0) are given by OP⃗=x1i^+y1j^+z1k^\vec{OP} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k} and OQ⃗=x2i^+y2j^+z2k^\vec{OQ} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k}.

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According to the triangle law of vector addition in △OPQ\triangle OPQ, we have OP⃗+PQ⃗=OQ⃗\vec{OP} + \vec{PQ} = \vec{OQ}.

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The vector PQ⃗\vec{PQ} joining the points PP and QQ is found by subtracting the position vector of the initial point from the position vector of the terminal point: PQ⃗=OQ⃗−OP⃗\vec{PQ} = \vec{OQ} - \vec{OP}.

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The magnitude of the vector PQ⃗\vec{PQ} represents the distance between the two points PP and QQ.

📐Formulae

PQ⃗=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}

∣PQ⃗∣=(x2−x1)2+(y2−y1)2+(z2−z1)2|\vec{PQ}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}

Direction Ratios of PQ⃗=(x2−x1),(y2−y1),(z2−z1)\text{Direction Ratios of } \vec{PQ} = (x_2 - x_1), (y_2 - y_1), (z_2 - z_1)

💡Examples

Problem 1:

Find the vector joining the points P(2,3,0)P(2, 3, 0) and Q(−1,−2,−4)Q(-1, -2, -4) directed from PP to QQ, and calculate its magnitude.

Solution:

Given points are P(x1,y1,z1)=(2,3,0)P(x_1, y_1, z_1) = (2, 3, 0) and Q(x2,y2,z2)=(−1,−2,−4)Q(x_2, y_2, z_2) = (-1, -2, -4). Using the formula PQ⃗=(x2−x1)i^+(y2−y1)j^+(z2−z1)k^\vec{PQ} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}, we compute the components:

x2−x1=−1−2=−3y2−y1=−2−3=−5z2−z1=−4−0=−4\begin{array}{r} x_2 - x_1 = -1 - 2 = -3 \\ y_2 - y_1 = -2 - 3 = -5 \\ z_2 - z_1 = -4 - 0 = -4 \end{array}

Thus, PQ⃗=−3i^−5j^−4k^\vec{PQ} = -3\hat{i} - 5\hat{j} - 4\hat{k}.

The magnitude is ∣PQ⃗∣=(−3)2+(−5)2+(−4)2=9+25+16=50=52|\vec{PQ}| = \sqrt{(-3)^2 + (-5)^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}.

Explanation:

First, identify the coordinates of the initial point PP and terminal point QQ. Subtract the coordinates of PP from QQ to get the vector components. Finally, use the square root of the sum of squares of these components to find the magnitude.

Problem 2:

If the vector joining A(1,2,k)A(1, 2, k) and B(4,6,15)B(4, 6, 15) has a magnitude of 1313 units, find the value of kk.

Solution:

The vector AB⃗=(4−1)i^+(6−2)j^+(15−k)k^=3i^+4j^+(15−k)k^\vec{AB} = (4-1)\hat{i} + (6-2)\hat{j} + (15-k)\hat{k} = 3\hat{i} + 4\hat{j} + (15-k)\hat{k}. Given ∣AB⃗∣=13|\vec{AB}| = 13, we have: 32+42+(15−k)2=13\sqrt{3^2 + 4^2 + (15-k)^2} = 13 Squaring both sides: 9+16+(15−k)2=1699 + 16 + (15-k)^2 = 169 25+(15−k)2=16925 + (15-k)^2 = 169 (15−k)2=144(15-k)^2 = 144 15−k=±1215-k = \pm 12 Case 1: 15−k=12⇒k=315-k = 12 \Rightarrow k = 3 Case 2: 15−k=−12⇒k=2715-k = -12 \Rightarrow k = 27 So, k=3k = 3 or k=27k = 27.

Explanation:

Form the vector AB⃗\vec{AB} in terms of kk, set its magnitude equal to 13, and solve the resulting quadratic equation for kk.