Consider two vectors a-2- b-3- and a third vector coplanar with a and b and also perpendicular to a with magnitude 14-5 units- Then the third vector is

Let third vector is c is coplanar with a and b is given by, c=xa+yb taking dot product with vector a both sides, ⇒c·a=x|a|^2+y(a·b) ⇒ 0=5x y [∵c⊥a] ⇒ 5x=y⋯(i) A ...

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Standard XII

Physics

Addition and Subtraction in Unit Vector Notation

Consider two ...

Question

Consider two vectors $\to a = 2^i -^j, \to b =^i + 3^j$ and a third vector coplanar with $\to a$ and $\to b$ and also perpendicular to $\to a$ with magnitude $14 \sqrt{5} units$ . Then the third vector is

14 (^i + 2^j)

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14 (^i - 2^j)

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- 14 (^i + 2^j)

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- 14 (^i - 2^j)

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Solution

The correct option is C $- 14 (^i + 2^j)$
Let third vector is $\to c$ is coplanar with $\to a$ and $\to b$ is given by,
$\to c = x \to a + y \to b$
taking dot product with vector $\to a$ both sides,
$\Rightarrow \to c \cdot \to a = x | \to a |^{2} + y (\to a \cdot \to b) \Rightarrow 0 = 5 x - y [∵ \to c ⊥ \to a] \Rightarrow 5 x = y \dots (i)$
Also,
$\to c = x (2^i -^j) + y (^i + 3^j) = (2 x + y)^i + (- x + 3 y)^j$
$\Rightarrow (2 x + y)^{2} + (- x + 3 y)^{2} = (14 \sqrt{5})^{2} [∵ | \to c | = 14 \sqrt{5}]$
using $(i); 49 x^{2} + 196 x^{2} = 980$
$\Rightarrow x^{2} = 4$
$\Rightarrow x = \pm 2; y = \pm 10$
$∴ \to c = 7 x^i + 14 x^j = \pm 14 (^i + 2^j)$

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Consider two vectors →a=2^i−^j,→b=^i+3^j and a third vector coplanar with →a and →b and also perpendicular to →a with magnitude 14√5 units. Then the third vector is

Consider two vectors $\to a = 2^i -^j, \to b =^i + 3^j$ and a third vector coplanar with $\to a$ and $\to b$ and also perpendicular to $\to a$ with magnitude $14 \sqrt{5} units$ . Then the third vector is