If θ is the angle between the vectors 4 ( ^ i - ^ k ) and ^ i + ^ j + ^ k , then (sinθ ++cosθ)equals to

Correct answer: C. 1 Let& → a = ^ i + ^ j + ^ k → b = 4 ( ^ i - ^ k ) Then, | → a | = 1 ^ 2 + 1 ^ 2 + 1 ^ 2 = 3 | → b | = 4 ^ 2 + 0 ^ 2 + (- 4)^ 2 =4 2 Now, cos θ = → a · → b | → a || → b | ⇒ cos θ = ( ^ i + ^ j + ^ k ) · (4 ^ i - 4 ^ k ) | → a || → b | ⇒ cos θ = 4 - 4 | → a || → b | =0 Hence, sin θ = 1 - ^ 2 θ =1 Therefore, sin θ + cos θ = 1

BITSAT 2018

Exam: BITSAT
Year: 2018
Subject: Mathematics
Chapter: Three Dimensional Geometry

BITSAT 2018 · Mathematics · Three Dimensional Geometry

If θ is the angle between the vectors 4 ( ^ i - ^ k ) and ^ i + ^ j...

\theta

is the angle between the vectors

4 \left(\overset{^}{i} - \overset{^}{k}\right) and \overset{^}{i} + \overset{^}{j} + \overset{^}{k},

then \l\left(sin\theta ++cos\theta\right)equals to

Options

A $0$
B $\dfrac{1}{2}$
C $1$
D $2$

Correct answer

C. $1$

Step-by-step solution

Let&

$\overset{→}{a} = \overset{^}{i} + \overset{^}{j} + \overset{^}{k}$

$\overset{→}{b} = 4 \left(\overset{^}{i} - \overset{^}{k}\right)Then,\left|\overset{→}{a}\right| = \sqrt{{1}^{2} + {1}^{2} + {1}^{2}} = \sqrt{3}\left|\overset{→}{b}\right| = \sqrt{{4}^{2} + {0}^{2} + \left(- 4\right)^{2}} =4\sqrt{2}Now,$ cos $\theta = \dfrac{\overset{→}{a} \cdot \overset{→}{b}}{\left|\overset{→}{a}\right|\left|\overset{→}{b}\right|}$ ⇒ cos $\theta = \dfrac{\left(\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right) \cdot \left(4\overset{^}{i} - 4\overset{^}{k}\right)}{\left|\overset{→}{a}\right|\left|\overset{→}{b}\right|}$ ⇒ cos $\theta = \dfrac{4 - 4}{\left|\overset{→}{a}\right|\left|\overset{→}{b}\right|} =0$

Hence, $sin$ \theta = $\sqrt{1 - \cos^{2}\theta} =1$

Therefore, $sin$ \theta + $cos$ \theta = $1$

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