If the projection of the vector \widehat i + 2\widehat j + \widehat k on the sum of the two vectors 2\widehat i + 4\widehat j - 5\widehat k and - \lambda \widehat i + 2\widehat j + 3\widehat k is 1, then \lambda is equal to \underline\hspace2cm .

Vector Algebra PYQ — JEE Main 2021 · 26 Aug · Shift 2 · Q41 with Solution

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Correct answer: 5

Let $\vec a=\hat i+2\hat j+\hat k,$ $\vec b=2\hat i+4\hat j-5\hat k,$ $\vec c=-\lambda \hat i+2\hat j+3\hat k.$

We are told that the projection of $\vec a$ on the sum of $\vec b$ and $\vec c$ is $1$ .

1. Find the sum vector

Let $\vec d=\vec b+\vec c.$ Then $\vec d=(2-\lambda)\hat i+(4+2)\hat j+(-5+3)\hat k$ $\vec d=(2-\lambda)\hat i+6\hat j-2\hat k.$

2. Use the formula for scalar projection

The scalar projection of $\vec a$ on $\vec d$ is $\text{proj}_{\vec d}(\vec a)=\frac{\vec a\cdot \vec d}{|\vec d|}.$ Given this is $1$ , $\frac{\vec a\cdot \vec d}{|\vec d|}=1.$

Now, $\vec a\cdot \vec d=1(2-\lambda)+2(6)+1(-2)=2-\lambda+12-2=12-\lambda.$

Also,

=\sqrt{(2-\lambda)^2+40}.$$ So, $$\frac{12-\lambda}{\sqrt{(2-\lambda)^2+40}}=1.$$ ## 3. Solve the equation Thus, $$12-\lambda=\sqrt{(2-\lambda)^2+40}.$$ Squaring both sides, $$(12-\lambda)^2=(2-\lambda)^2+40.$$ Expand: $$\lambda^2-24\lambda+144=\lambda^2-4\lambda+4+40.$$ $$\lambda^2-24\lambda+144=\lambda^2-4\lambda+44.$$ Cancel $\lambda^2$: $$-24\lambda+144=-4\lambda+44.$$ $$100=20\lambda.$$ $$\lambda=5.$$ ## 4. Check For $\lambda=5$, $$\vec d=(-3)\hat i+6\hat j-2\hat k.$$ Then $$\vec a\cdot \vec d=-3+12-2=7,$$ $$|\vec d|=\sqrt{9+36+4}=\sqrt{49}=7.$$ Hence, $$\frac{\vec a\cdot \vec d}{|\vec d|}=\frac{7}{7}=1,$$ which satisfies the condition. Therefore, the required value is $$\boxed{5}.$$

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2. Use the formula for scalar projection

The scalar projection of

\vec a

\vec d

\text{proj}_{\vec d}(\vec a)=\frac{\vec a\cdot \vec d}{|\vec d|}.

Given this is

1

\frac{\vec a\cdot \vec d}{|\vec d|}=1.

Now,

\vec a\cdot \vec d=1(2-\lambda)+2(6)+1(-2)=2-\lambda+12-2=12-\lambda.

Also,

=\sqrt{(2-\lambda)^2+40}.$$ So, $$\frac{12-\lambda}{\sqrt{(2-\lambda)^2+40}}=1.$$ ## 3. Solve the equation Thus, $$12-\lambda=\sqrt{(2-\lambda)^2+40}.$$ Squaring both sides, $$(12-\lambda)^2=(2-\lambda)^2+40.$$ Expand: $$\lambda^2-24\lambda+144=\lambda^2-4\lambda+4+40.$$ $$\lambda^2-24\lambda+144=\lambda^2-4\lambda+44.$$ Cancel $\lambda^2$: $$-24\lambda+144=-4\lambda+44.$$ $$100=20\lambda.$$ $$\lambda=5.$$ ## 4. Check For $\lambda=5$, $$\vec d=(-3)\hat i+6\hat j-2\hat k.$$ Then $$\vec a\cdot \vec d=-3+12-2=7,$$ $$|\vec d|=\sqrt{9+36+4}=\sqrt{49}=7.$$ Hence, $$\frac{\vec a\cdot \vec d}{|\vec d|}=\frac{7}{7}=1,$$ which satisfies the condition. Therefore, the required value is $$\boxed{5}.$$

Vector Algebra question

Vector Algebra question

1. Find the sum vector

2. Use the formula for scalar projection

More from Vector Algebra

Vector Algebra question

Vector Algebra question

1. Find the sum vector

2. Use the formula for scalar projection

More from Vector Algebra